PROBLEM LINK:
Author: Takuki Kurokawa
Tester: Aryan Choudhary
DIFFICULTY:
MEDIUM-HARD
PREREQUISITES:
Dynamic Programming.
PROBLEM:
You are given a positive integer K and a tree with N vertices, rooted at vertex 1.
For all integer i (2 \leq i \leq N), P_i is a parent of vertex i.
Let’s call an array A consisting of N positive integers brilliant if the following constraints are met;
-
A_{P_i}\bmod A_i = 0, for all integer i (2 \leq i \leq N).
-
\prod_{i=1}^N A_i \leq K.
Compute remainder modulo 998244353 of the number of possible brilliant arrays.
QUICK EXPLANATION (HINTS):
Hint 1
What if the product of array is restricted to power of 2?
Hint 2
Use tree DP and solve Hint 1.
Hint 3
Solve Hint 1 for each prime and combine them. Can you solve with K = 100000?
Hint 4
Hint 3 can be solved using DP. The key should be K / (product of A), not product of A. By the way, when K = 100, can A[2] be 11?
Hint 5
A[1] is the only one that can be multiple of prime more than sqrt(K). If you can exclude these primes, this DP can be computed with K = 1e8. But how?
Hint 6
Add new constraint: A[1] = lcm(A[2], A[3], …, A[N]).
Hint 7
With the constraint of Hint 6, you can compute this DP using two arrays.
Hint 8
With the constraint of Hint 6, there are values whose product of A cannot be. For example, 2, 12, 42.
Hint 9
If product of A has a prime factor p, it also should be divided by square of p. These number are called powerful number.
Hint 10
Let M be the number of powerful number not more than K. With K = 1e12, M is about 2e6 which is small enough.
Hint 11
Use associative array instead of array to compute DP. The number of transitions of DP will be at most M.
EXPLANATION:
Consider the case that each element of A can be expressed as power of a prime p, so A_i = p^j.
Let dp^1_{x,y,z} be the number of arrays A which satisfy
-
A_x = p^y
-
\prod_{i \in \textrm{subtree}(x)} A_i = p^z
You can calculate this DP using DFS on tree. Since y, z \leq \log K, the time complexity is O(N \log^4 K) (setter’s solution). You can improve this to O(N \log^3 K) using prefix sum and inclusive-exclusive principle (tester’s solution).
Next, let’s combine primes. There are O(K/\log K) primes less than or equals to K, which is too many.
Notice that if p>\sqrt K, A_1 is the only element which can be multiple of p. Let’s add a new constraint: A_1 = \textrm{lcm}_{i\neq 1}(A_i). In other words, A_1 will be the smallest possible value. With this constraint, you can put off primes more than \sqrt K. You can multiply to A_1 between 1 and \lfloor K/\prod_{i\neq 1}A_i\rfloor later.
Let F_z be the number of arrays A which satisfy
-
\prod_{i=1}^N A_i = p^z
-
A_1 = \textrm{lcm}_{i\neq 1}(A_i) = \max_{i\neq 1}(A_i)
Let dp^2_{x,y} be number of arrays A which satisfy
-
\prod_{i=1}^N A_i has only prime factors between P_1 and P_x. (P_i denotes the i-th prime.)
-
\frac{K}{\prod_{i=1}^N A_i} = y
-
A_1=\textrm{lcm}_{i\neq 1}(A_i)
This DP’s transitions are dp^2_{x+1,\lfloor y/ P_x^i\rfloor}+=dp^2_{x,y}\times F_i. The initial value is dp^2_{0,K}=1.
There are O(\sqrt K / \log K) different primes so O(\sqrt K) both primes and perfect power of them. O(\sqrt K) keys in dp^2, therefore using two arrays whose boundary is \sqrt K, this DP runs in O(K).
What happen when you use associative array like std::map ? Let M be the number of positive integers which is not more than K and only have prime factors whose exponent are at least 2. Since F_1=0, the number of DP’s transitions is at most M. It can be shown that M=O(\sqrt K) (details are below), so this DP runs in O(\sqrt K\log K).
You can also use DFS and enumerate all possible these M values. This solution runs in O(\sqrt K) (tester’s solution).
Finally, consider vertex 1. You can multiply to A_1 between 1 to x, so you can just sum up the multiply of each key and value of dp^2.
The overall time complexity is O(N\log^4K +\sqrt K \log K).
The number X is called powerful number if for each prime factor p of X, p^2 divides X. X can be expressed as X=A^2B^3 with two positive integer A,B. Let’s count M', the number of pair (A,B) which satisfies A^2B^3\leq K.
From the discussion above, you can get
M\leq M'= K^{\frac{1}{3}}\sum_{i=1}^{\sqrt K}i^{-\frac{2}{3}}
The sigma part can be evaluated as
\int_{1}^{\sqrt K}x^{-\frac{2}{3}}\ dx=\left[3x^\frac{1}{3} \right]_1^{\sqrt K}\approx K^\frac{1}{6}
So M=O(\sqrt K).
ALTERNATE EXPLANATION:
dp^1 and F are multiplicative function.
Prefix sum of multiplicative function can be computed in sublinear time.
SOLUTIONS:
Setter's Solution
#include <bits/stdc++.h>
using namespace std;
int main() {
const long long mod = 998244353;
const int LOG_K = 42;
const int SQRT_K = (int) 1e6 + 16;
int n;
long long k;
cin >> n >> k;
vector<int> p(n, -1);
for (int i = 1; i < n; i++) {
cin >> p[i];
p[i]--;
}
vector<vector<int>> g(n);
for (int i = 1; i < n; i++) {
g[p[i]].emplace_back(i);
}
vector dp1(n, vector(LOG_K, vector<long long>(LOG_K)));
function<void(int)> dfs = [&](int v) {
dp1[v][0][0] = 1;
for (int to : g[v]) {
dfs(to);
vector new_dp1(LOG_K, vector<long long>(LOG_K));
for (int i = 0; i < LOG_K; i++) {
for (int j = i; j < LOG_K; j++) {
for (int ni = 0; ni < LOG_K; ni++) {
for (int nj = ni; j + nj < LOG_K; nj++) {
new_dp1[max(i, ni)][j + nj] = (new_dp1[max(i, ni)][j + nj] + dp1[v][i][j] * dp1[to][ni][nj]) % mod;
}
}
}
}
swap(dp1[v], new_dp1);
}
vector new_dp1(LOG_K, vector<long long>(LOG_K));
for (int i = 0; i < LOG_K; i++) {
for (int j = i; j < LOG_K; j++) {
for (int ni = i; j + ni < LOG_K; ni++) {
if (v == 0 && ni != i) {
break;
}
new_dp1[ni][j + ni] += dp1[v][i][j];
if (new_dp1[ni][j + ni] >= mod) {
new_dp1[ni][j + ni] -= mod;
}
}
}
}
swap(dp1[v], new_dp1);
};
dfs(0);
vector<long long> f(LOG_K);
for (int i = 0; i < LOG_K; i++) {
for (int j = 0; j < LOG_K; j++) {
f[j] += dp1[0][i][j];
if (f[j] >= mod) {
f[j] -= mod;
}
}
}
vector<bool> is_prime(SQRT_K, true);
is_prime[0] = is_prime[1] = false;
map<long long, long long> dp2;
dp2[k] = 1;
for (int x = 2; x <= k / x; x++) {
if (!is_prime[x]) {
continue;
}
for (int y = x * 2; y < SQRT_K; y += x) {
is_prime[y] = false;
}
vector<long long> pows(1, 1);
while (pows.back() <= k / x) {
pows.emplace_back(pows.back() * x);
}
for (auto iter = dp2.lower_bound(pows[2]); iter != dp2.end(); iter++) {
auto [remain, value] = *iter;
for (int i = 2; i < (int) pows.size(); i++) {
if (remain < pows[i]) {
break;
}
dp2[remain / pows[i]] = (dp2[remain / pows[i]] + value * f[i]) % mod;
}
}
}
long long ans = 0;
for (auto [remain, value] : dp2) {
ans = (ans + remain % mod * value) % mod;
}
cout << ans << '\n';
return 0;
}
Tester's Solution
/*
Compete against Yourself.
Author - Aryan (@aryanc403)
*/
/*
Credits -
Atcoder library - https://atcoder.github.io/ac-library/production/document_en/ (namespace atcoder)
Github source code of library - https://github.com/atcoder/ac-library/tree/master/atcoder
*/
#ifdef ARYANC403
#include <header.h>
#else
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define dbg(args...) 42;
#define endl "\n"
#endif
// y_combinator from @neal template https://codeforces.com/contest/1553/submission/123849801
// http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html
template<class Fun> class y_combinator_result {
Fun fun_;
public:
template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }
using namespace std;
#define fo(i,n) for(i=0;i<(n);++i)
#define repA(i,j,n) for(i=(j);i<=(n);++i)
#define repD(i,j,n) for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;
template <class T>
using ordered_set = __gnu_pbds::tree<T,__gnu_pbds::null_type,less<T>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>;
// X.find_by_order(k) return kth element. 0 indexed.
// X.order_of_key(k) returns count of elements strictly less than k.
const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}
const lli INF = 0xFFFFFFFFFFFFFFFLL;
const lli SEED=chrono::steady_clock::now().time_since_epoch().count();
mt19937 rng(SEED);
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}
class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{ return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y )); }};
void add( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt==m.end()) m.insert({x,cnt});
else jt->Y+=cnt;
}
void del( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt->Y<=cnt) m.erase(jt);
else jt->Y-=cnt;
}
bool cmp(const ii &a,const ii &b)
{
return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}
const lli mod = 1000000007L;
// const lli maxN = 1000000007L;
#include <cassert>
#include <numeric>
#include <type_traits>
#ifdef _MSC_VER
#include <intrin.h>
#endif
#include <utility>
#ifdef _MSC_VER
#include <intrin.h>
#endif
namespace atcoder {
namespace internal {
constexpr long long safe_mod(long long x, long long m) {
x %= m;
if (x < 0) x += m;
return x;
}
struct barrett {
unsigned int _m;
unsigned long long im;
explicit barrett(unsigned int m) : _m(m), im((unsigned long long)(-1) / m + 1) {}
unsigned int umod() const { return _m; }
unsigned int mul(unsigned int a, unsigned int b) const {
unsigned long long z = a;
z *= b;
#ifdef _MSC_VER
unsigned long long x;
_umul128(z, im, &x);
#else
unsigned long long x =
(unsigned long long)(((unsigned __int128)(z)*im) >> 64);
#endif
unsigned int v = (unsigned int)(z - x * _m);
if (_m <= v) v += _m;
return v;
}
};
constexpr long long pow_mod_constexpr(long long x, long long n, int m) {
if (m == 1) return 0;
unsigned int _m = (unsigned int)(m);
unsigned long long r = 1;
unsigned long long y = safe_mod(x, m);
while (n) {
if (n & 1) r = (r * y) % _m;
y = (y * y) % _m;
n >>= 1;
}
return r;
}
constexpr bool is_prime_constexpr(int n) {
if (n <= 1) return false;
if (n == 2 || n == 7 || n == 61) return true;
if (n % 2 == 0) return false;
long long d = n - 1;
while (d % 2 == 0) d /= 2;
constexpr long long bases[3] = {2, 7, 61};
for (long long a : bases) {
long long t = d;
long long y = pow_mod_constexpr(a, t, n);
while (t != n - 1 && y != 1 && y != n - 1) {
y = y * y % n;
t <<= 1;
}
if (y != n - 1 && t % 2 == 0) {
return false;
}
}
return true;
}
template <int n> constexpr bool is_prime = is_prime_constexpr(n);
constexpr std::pair<long long, long long> inv_gcd(long long a, long long b) {
a = safe_mod(a, b);
if (a == 0) return {b, 0};
long long s = b, t = a;
long long m0 = 0, m1 = 1;
while (t) {
long long u = s / t;
s -= t * u;
m0 -= m1 * u; // |m1 * u| <= |m1| * s <= b
auto tmp = s;
s = t;
t = tmp;
tmp = m0;
m0 = m1;
m1 = tmp;
}
if (m0 < 0) m0 += b / s;
return {s, m0};
}
constexpr int primitive_root_constexpr(int m) {
if (m == 2) return 1;
if (m == 167772161) return 3;
if (m == 469762049) return 3;
if (m == 754974721) return 11;
if (m == 998244353) return 3;
int divs[20] = {};
divs[0] = 2;
int cnt = 1;
int x = (m - 1) / 2;
while (x % 2 == 0) x /= 2;
for (int i = 3; (long long)(i)*i <= x; i += 2) {
if (x % i == 0) {
divs[cnt++] = i;
while (x % i == 0) {
x /= i;
}
}
}
if (x > 1) {
divs[cnt++] = x;
}
for (int g = 2;; g++) {
bool ok = true;
for (int i = 0; i < cnt; i++) {
if (pow_mod_constexpr(g, (m - 1) / divs[i], m) == 1) {
ok = false;
break;
}
}
if (ok) return g;
}
}
template <int m> constexpr int primitive_root = primitive_root_constexpr(m);
unsigned long long floor_sum_unsigned(unsigned long long n,
unsigned long long m,
unsigned long long a,
unsigned long long b) {
unsigned long long ans = 0;
while (true) {
if (a >= m) {
ans += n * (n - 1) / 2 * (a / m);
a %= m;
}
if (b >= m) {
ans += n * (b / m);
b %= m;
}
unsigned long long y_max = a * n + b;
if (y_max < m) break;
n = (unsigned long long)(y_max / m);
b = (unsigned long long)(y_max % m);
std::swap(m, a);
}
return ans;
}
} // namespace internal
} // namespace atcoder
#include <cassert>
#include <numeric>
#include <type_traits>
namespace atcoder {
namespace internal {
#ifndef _MSC_VER
template <class T>
using is_signed_int128 =
typename std::conditional<std::is_same<T, __int128_t>::value ||
std::is_same<T, __int128>::value,
std::true_type,
std::false_type>::type;
template <class T>
using is_unsigned_int128 =
typename std::conditional<std::is_same<T, __uint128_t>::value ||
std::is_same<T, unsigned __int128>::value,
std::true_type,
std::false_type>::type;
template <class T>
using make_unsigned_int128 =
typename std::conditional<std::is_same<T, __int128_t>::value,
__uint128_t,
unsigned __int128>;
template <class T>
using is_integral = typename std::conditional<std::is_integral<T>::value ||
is_signed_int128<T>::value ||
is_unsigned_int128<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using is_signed_int = typename std::conditional<(is_integral<T>::value &&
std::is_signed<T>::value) ||
is_signed_int128<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using is_unsigned_int =
typename std::conditional<(is_integral<T>::value &&
std::is_unsigned<T>::value) ||
is_unsigned_int128<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using to_unsigned = typename std::conditional<
is_signed_int128<T>::value,
make_unsigned_int128<T>,
typename std::conditional<std::is_signed<T>::value,
std::make_unsigned<T>,
std::common_type<T>>::type>::type;
#else
template <class T> using is_integral = typename std::is_integral<T>;
template <class T>
using is_signed_int =
typename std::conditional<is_integral<T>::value && std::is_signed<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using is_unsigned_int =
typename std::conditional<is_integral<T>::value &&
std::is_unsigned<T>::value,
std::true_type,
std::false_type>::type;
template <class T>
using to_unsigned = typename std::conditional<is_signed_int<T>::value,
std::make_unsigned<T>,
std::common_type<T>>::type;
#endif
template <class T>
using is_signed_int_t = std::enable_if_t<is_signed_int<T>::value>;
template <class T>
using is_unsigned_int_t = std::enable_if_t<is_unsigned_int<T>::value>;
template <class T> using to_unsigned_t = typename to_unsigned<T>::type;
} // namespace internal
} // namespace atcoder
namespace atcoder {
namespace internal {
struct modint_base {};
struct static_modint_base : modint_base {};
template <class T> using is_modint = std::is_base_of<modint_base, T>;
template <class T> using is_modint_t = std::enable_if_t<is_modint<T>::value>;
} // namespace internal
template <int m, std::enable_if_t<(1 <= m)>* = nullptr>
struct static_modint : internal::static_modint_base {
using mint = static_modint;
public:
static constexpr int mod() { return m; }
static mint raw(int v) {
mint x;
x._v = v;
return x;
}
static_modint() : _v(0) {}
template <class T, internal::is_signed_int_t<T>* = nullptr>
static_modint(T v) {
long long x = (long long)(v % (long long)(umod()));
if (x < 0) x += umod();
_v = (unsigned int)(x);
}
template <class T, internal::is_unsigned_int_t<T>* = nullptr>
static_modint(T v) {
_v = (unsigned int)(v % umod());
}
unsigned int val() const { return _v; }
mint& operator++() {
_v++;
if (_v == umod()) _v = 0;
return *this;
}
mint& operator--() {
if (_v == 0) _v = umod();
_v--;
return *this;
}
mint operator++(int) {
mint result = *this;
++*this;
return result;
}
mint operator--(int) {
mint result = *this;
--*this;
return result;
}
mint& operator+=(const mint& rhs) {
_v += rhs._v;
if (_v >= umod()) _v -= umod();
return *this;
}
mint& operator-=(const mint& rhs) {
_v -= rhs._v;
if (_v >= umod()) _v += umod();
return *this;
}
mint& operator*=(const mint& rhs) {
unsigned long long z = _v;
z *= rhs._v;
_v = (unsigned int)(z % umod());
return *this;
}
mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }
mint operator+() const { return *this; }
mint operator-() const { return mint() - *this; }
mint pow(long long n) const {
assert(0 <= n);
mint x = *this, r = 1;
while (n) {
if (n & 1) r *= x;
x *= x;
n >>= 1;
}
return r;
}
mint inv() const {
if (prime) {
assert(_v);
return pow(umod() - 2);
} else {
auto eg = internal::inv_gcd(_v, m);
assert(eg.first == 1);
return eg.second;
}
}
friend mint operator+(const mint& lhs, const mint& rhs) {
return mint(lhs) += rhs;
}
friend mint operator-(const mint& lhs, const mint& rhs) {
return mint(lhs) -= rhs;
}
friend mint operator*(const mint& lhs, const mint& rhs) {
return mint(lhs) *= rhs;
}
friend mint operator/(const mint& lhs, const mint& rhs) {
return mint(lhs) /= rhs;
}
friend bool operator==(const mint& lhs, const mint& rhs) {
return lhs._v == rhs._v;
}
friend bool operator!=(const mint& lhs, const mint& rhs) {
return lhs._v != rhs._v;
}
private:
unsigned int _v;
static constexpr unsigned int umod() { return m; }
static constexpr bool prime = internal::is_prime<m>;
};
template <int id> struct dynamic_modint : internal::modint_base {
using mint = dynamic_modint;
public:
static int mod() { return (int)(bt.umod()); }
static void set_mod(int m) {
assert(1 <= m);
bt = internal::barrett(m);
}
static mint raw(int v) {
mint x;
x._v = v;
return x;
}
dynamic_modint() : _v(0) {}
template <class T, internal::is_signed_int_t<T>* = nullptr>
dynamic_modint(T v) {
long long x = (long long)(v % (long long)(mod()));
if (x < 0) x += mod();
_v = (unsigned int)(x);
}
template <class T, internal::is_unsigned_int_t<T>* = nullptr>
dynamic_modint(T v) {
_v = (unsigned int)(v % mod());
}
unsigned int val() const { return _v; }
mint& operator++() {
_v++;
if (_v == umod()) _v = 0;
return *this;
}
mint& operator--() {
if (_v == 0) _v = umod();
_v--;
return *this;
}
mint operator++(int) {
mint result = *this;
++*this;
return result;
}
mint operator--(int) {
mint result = *this;
--*this;
return result;
}
mint& operator+=(const mint& rhs) {
_v += rhs._v;
if (_v >= umod()) _v -= umod();
return *this;
}
mint& operator-=(const mint& rhs) {
_v += mod() - rhs._v;
if (_v >= umod()) _v -= umod();
return *this;
}
mint& operator*=(const mint& rhs) {
_v = bt.mul(_v, rhs._v);
return *this;
}
mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }
mint operator+() const { return *this; }
mint operator-() const { return mint() - *this; }
mint pow(long long n) const {
assert(0 <= n);
mint x = *this, r = 1;
while (n) {
if (n & 1) r *= x;
x *= x;
n >>= 1;
}
return r;
}
mint inv() const {
auto eg = internal::inv_gcd(_v, mod());
assert(eg.first == 1);
return eg.second;
}
friend mint operator+(const mint& lhs, const mint& rhs) {
return mint(lhs) += rhs;
}
friend mint operator-(const mint& lhs, const mint& rhs) {
return mint(lhs) -= rhs;
}
friend mint operator*(const mint& lhs, const mint& rhs) {
return mint(lhs) *= rhs;
}
friend mint operator/(const mint& lhs, const mint& rhs) {
return mint(lhs) /= rhs;
}
friend bool operator==(const mint& lhs, const mint& rhs) {
return lhs._v == rhs._v;
}
friend bool operator!=(const mint& lhs, const mint& rhs) {
return lhs._v != rhs._v;
}
private:
unsigned int _v;
static internal::barrett bt;
static unsigned int umod() { return bt.umod(); }
};
template <int id> internal::barrett dynamic_modint<id>::bt(998244353);
using modint998244353 = static_modint<998244353>;
using modint1000000007 = static_modint<1000000007>;
using modint = dynamic_modint<-1>;
namespace internal {
template <class T>
using is_static_modint = std::is_base_of<internal::static_modint_base, T>;
template <class T>
using is_static_modint_t = std::enable_if_t<is_static_modint<T>::value>;
template <class> struct is_dynamic_modint : public std::false_type {};
template <int id>
struct is_dynamic_modint<dynamic_modint<id>> : public std::true_type {};
template <class T>
using is_dynamic_modint_t = std::enable_if_t<is_dynamic_modint<T>::value>;
} // namespace internal
} // namespace atcoder
using namespace atcoder;
using mint = modint998244353;
//using mint = modint1000000007;
using vm = vector<mint>;
std::ostream& operator << (std::ostream& out, const mint& rhs) {
return out<<rhs.val();
}
const int N = 43;
using dpType = array<array<mint,N>,N>;
const int LEN = 1<<20;
using bt = bitset<LEN>;
void primesUptoSqrt(lli n,vi &primes){
const lli sq=(lli)(sqrt(n+1))+10;
vector<bool> vis(sq+1);
for(lli i=2;i*i<=n;++i){
if(vis[i])
continue;
primes.pb(i);
for(lli j=i*i;j*j<=n;j+=i)
vis[j]=true;
}
dbg(primes);
}
void primesUptoN(const lli n,vi &primes){
primesUptoSqrt(n,primes);
bt vis;
vis[0]=vis[1]=1;
for(lli l=0;l<n;l+=LEN){
for(const auto &p:primes){
if(p>LEN)
break;
for(lli i=p-(l%p);i<LEN;i+=p)
vis[i]=1;
}
for(lli i=1;i<LEN;++i){
if(!vis[i])
primes.pb(i);
}
vis.reset();
}
}
int main(void) {
ios_base::sync_with_stdio(false);cin.tie(NULL);
// freopen("txt.in", "r", stdin);
// freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
// cin>>T;while(T--)
{
lli n,k;
cin>>n>>k;
cerr<<"k "<<k<<endl;
vector<vi> e(n);
for(int i=1;i<n;++i){
int p;
cin>>p;
p--;
e[p].pb(i);
}
auto merge=[&](dpType &a,dpType b){
dpType res;
for(int am=0;am<N;++am)
for(int ai=0;ai<N;++ai)
for(int bi=0;bi+ai<N;++bi)
res[am][ai+bi]+=a[am][ai]*b[am][bi];
return res;
};
auto addValue=[&](dpType a,bool fl){
dpType res;
for(int am=N-1;am>0;--am)
for(int ai=0;ai<N;++ai)
a[am][ai]-=a[am-1][ai];
if(fl)
return a;
for(int pm=0;pm<N;++pm)
for(int ai=0;ai+pm<N;++ai)
for(int am=0;am<=pm&&am<=ai;++am)
res[pm][ai+pm]+=a[am][ai];
for(int ai=0;ai<N;++ai)
for(int am=1;am<N;++am){
res[am][ai]+=res[am-1][ai];
}
return res;
};
const auto dpWithMax=y_combinator([&](const auto &self,const int u)->dpType{
dpType cur;
for(int i=0;i<N;++i)
cur[i][0]=1;
for(auto x:e[u])
cur=merge(cur,self(x));
cur=addValue(cur,u==0);
return cur;
})(0);
vm ways(N-2);
for(int at=2;at<N;++at)
for(int am=0;2*am<=at;++am)
ways[at-2]+=dpWithMax[am][at-am];
vi primes;
primesUptoN(LEN,primes);
mint ans=k;
// lli dfsCnt=0;
y_combinator([&](const auto &self,lli val,lli pidx,mint fac)->void{
// dfsCnt++;
if(pidx==sz(primes)||primes[pidx]*primes[pidx]>val)
return;
self(val,pidx+1,fac);
const lli p=primes[pidx];
val/=p;
val/=p;
for(int i=0;val;++i,val/=p){
ans+=val*ways[i]*fac;
self(val,pidx+1,fac*ways[i]);
}
})(k,0,1);
// cerr<<"dfsCnt : "<<dfsCnt<<endl;
cout<<ans<<endl;
} aryanc403();
return 0;
}