PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4
Author: pols_agyi_pols
Tester: kingmessi
Editorialist: iceknight1093
DIFFICULTY:
easy-med
PREREQUISITES:
Combinatorics, inclusion-exclusion
PROBLEM:
For two permutations A and B of the same size, we define f(A, B) as follows.
- Let i = 1 initially.
- For each j from 1 to N in order,
- If A_i = B_j, add 1 to F(A, B).
- Otherwise, increment i by 1.
You’re given N and K, as well as a partially filled permutation P. Count the number of permutation pairs (A, B) of length N such that F(A, B) = K, and A matches P at all positions where P is filled.
EXPLANATION:
Let’s fix a single permutation A, and try to count the number of B such that F(A, B) = K.
Observe that if B is fixed, the process can be described by a binary string of length N: with each character being 0 if i was incremented during the process (corresponds to a removal in the original setup), and 1 otherwise (corresponds to a skip in the original setup).
F(A, B) of course equals the number of ones in this binary string.
Now, let S be a binary string of length N corresponding to this process.
Observe that if S_j = 1, then S_{j+1} = 0 must hold (unless j = N).
This is because S_j = 1 means B_j = A_i occurred; and the next comparison would’ve been A_i against B_{j+1}, which cannot be equal since B is also a permutation.
So, for F(A, B) = K, S should be in one of two forms:
- Contain exactly K ‘blocks’ of 10, and all other characters are 0; or
- Contain exactly K-1 ‘blocks’ of 10, the last character is 1, and all other characters are 0.
We’ll deal with just the first type for now; the second can be handled similarly.
Let’s fix a binary string S of this form (i.e. there are K blocks of 10 and everything else is 0).
How many permutations B correspond to this string?
Suppose the ones in S are at indices i_1 \lt i_2 \lt\cdots\lt i_K.
Note that i_j+1 \lt i_{j+1} holds here, since each 1 must be followed by a 0.
Now, looking at how the process runs:
- For each 1 \leq i \lt i_1, A_i \neq B_i must hold.
- We then have A_{i_1} = B_{i_1}.
- Next, we want B_{i_1 + 1} \neq A_{i_1} to hold; but this will always be true no matter what B_{i_1+1} is chosen to be.
- Then, for each i_1 \lt i \lt i_2, A_{i-1} \neq B_{i} must hold.
- Following that, A_{i_2-1} = B_{i_2}, and so on.
In general, we see that:
- For each 1 \leq j \leq K, B_{i_j} should be equal to A_{i_j - (j-1)}.
- B_{i_j + 1} can be anything at all without affecting any conditions.
- For every other element, it will be compared against exactly one element of A, and should not equal that single element.
From the first point, some positions of B are fixed - let’s fix them and forget about them.
There are N-K positions remaining.
Of them, K are “free” and can contain anything, while the remaining N-2K each have a specific value they aren’t allowed to take (and two different indices do not have the same element forbidden).
We see that this is pretty similar to a derangement: in fact, if K = 0 this would be exactly a derangement!
In particular, we can adapt methods of counting derangements to this task.
For example, here’s one way to do it using inclusion-exclusion:
- Fix i (0 \leq i \leq N-2K), the number of indices that violate their constraints.
- There are \binom{N-2K}{i} ways to choose such a subset, and (N-K-i)! ways to arrange the remaining elements.
- Add or subtract this product appropriately, depending on the parity of i.
So, for a fixed permutation A and binary string S, we’re able to count the number of B such that F(A, B) = K.
(Note that the case where the last element of S equals 1 wasn’t explicitly worked out, but is similar).
However, we can make a crucial observation: the count didn’t really depend on either A or S at all! It depends only on N and K.
This means, for a fixed A, the total number of valid B simply equals the answer for a single S, multiplied by the number of valid binary strings S.
The number of valid binary strings S is easy to count using stars-and-bars.
Finally, once we know the answer for a single A, we can multiply this by the number of valid A to obtain the overall answer.
The number of valid A is easy: count the number of empty positions, and take its factorial.
TIME COMPLEXITY:
\mathcal{O}(N\log\text{MOD}) per testcase.
CODE:
Author's code (C++)
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define ll long long
using namespace std;
#define ll long long
const ll N=1e6+1;
ll factorialNumInverse[N + 1];
ll naturalNumInverse[N + 1];
ll factorial[N + 1];
ll Binomial(ll N, ll R, ll p)
{
ll ans = ((factorial[N] * factorialNumInverse[R])
% p * factorialNumInverse[N - R])
% p;
return ans;
}
void InverseofNumber(ll p)
{
naturalNumInverse[0] = naturalNumInverse[1] = 1;
for (int i = 2; i <= N; i++)
naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
}
void InverseofFactorial(ll p)
{
factorialNumInverse[0] = factorialNumInverse[1] = 1;
for (int i = 2; i <= N; i++)
factorialNumInverse[i] = (naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
}
void FACTORIAL(ll p)
{
factorial[0] = 1;
for (int i = 1; i <= N; i++) {
factorial[i] = (factorial[i - 1] * i) % p;
}
}
void start()
{
ll p=1000000007;
InverseofNumber(p);
InverseofFactorial(p);
FACTORIAL(p);
}
ll solve(ll n,ll k){
ll ans=0;
ll cnt;
for(int p=0;p<=k;p++){
if(p%2){
cnt=(Binomial(k,p,1000000007)*factorial[n-p]);
cnt%=1000000007;
ans-=cnt;
}else{
cnt=(Binomial(k,p,1000000007)*factorial[n-p]);
cnt%=1000000007;
ans+=cnt;
}
ans+=1000000007;
}
ans%=1000000007;
return ans;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
start();
ll kitne_cases_hain;
kitne_cases_hain=1;
cin>>kitne_cases_hain;
while(kitne_cases_hain--){
ll n,k;
cin>>n>>k;
ll x,y;
y=0;
for(int i=0;i<n;i++){
cin>>x;
if(x==-1){
y++;
}
}
if(n==1 && k==0){
cout<<"0\n";
continue;
}
if(2*k>(n+1)){
cout<<"0\n";
continue;
}
ll ans;
if((2*k-1)==n){
ans=factorial[n-k];ans*=factorial[y];
ans%=1000000007;
cout<<ans<<"\n";
continue;
}
if(k==0){
ans=solve(n,n);
}else{
ans=(solve(n-k,n-2*k)*Binomial(n-k,k,1000000007))%1000000007;
ans+=(solve(n-k,n-2*k+1)*Binomial(n-k,k-1,1000000007))%1000000007;ans%=1000000007;
}
ans*=factorial[y];ans%=1000000007;
cout<<ans<<"\n";
}
return 0;
}
Tester's code (C++)
#include<bits/stdc++.h>
#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 long long y = x * _m;
return (unsigned int)(z - y + (z < y ? _m : 0));
}
};
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
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define int long long
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define repin rep(i,0,n)
#define di(a) int a;cin>>a;
#define precise(i) cout<<fixed<<setprecision(i)
#define vi vector<int>
#define si set<int>
#define mii map<int,int>
#define take(a,n) for(int j=0;j<n;j++) cin>>a[j];
#define give(a,n) for(int j=0;j<n;j++) cout<<a[j]<<' ';
#define vpii vector<pair<int,int>>
#define sis string s;
#define sin string s;cin>>s;
#define db double
#define be(x) x.begin(),x.end()
#define pii pair<int,int>
#define pb push_back
#define pob pop_back
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#define bpc(x) __builtin_popcountll(x)
#define btz(x) __builtin_ctz(x)
using namespace std;
using namespace atcoder;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<pair<int, int>, null_type,less<pair<int, int> >, rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
const long long INF=1e18;
const long long M=1e9+7;
const long long MM=998244353;
using mint = static_modint<M>;
int power( int N, int M){
int power = N, sum = 1;
if(N == 0) sum = 0;
while(M > 0){if((M & 1) == 1){sum *= power;}
power = power * power;M = M >> 1;}
return sum;
}
struct input_checker {
string buffer;
int pos;
const string all = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
const string number = "0123456789";
const string upper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const string lower = "abcdefghijklmnopqrstuvwxyz";
input_checker() {
pos = 0;
while (true) {
int c = cin.get();
if (c == -1) {
break;
}
buffer.push_back((char) c);
}
}
int nextDelimiter() {
int now = pos;
while (now < (int) buffer.size() && !isspace(buffer[now])) {
now++;
}
return now;
}
string readOne() {
assert(pos < (int) buffer.size());
int nxt = nextDelimiter();
string res;
while (pos < nxt) {
res += buffer[pos];
pos++;
}
return res;
}
string readString(int minl, int maxl, const string &pattern = "") {
assert(minl <= maxl);
string res = readOne();
assert(minl <= (int) res.size());
assert((int) res.size() <= maxl);
for (int i = 0; i < (int) res.size(); i++) {
assert(pattern.empty() || pattern.find(res[i]) != string::npos);
}
return res;
}
int readInt(int minv, int maxv) {
assert(minv <= maxv);
int res = stoi(readOne());
assert(minv <= res);
assert(res <= maxv);
return res;
}
long long readLong(long long minv, long long maxv) {
assert(minv <= maxv);
long long res = stoll(readOne());
assert(minv <= res);
assert(res <= maxv);
return res;
}
auto readInts(int n, int minv, int maxv) {
assert(n >= 0);
vector<int> v(n);
for (int i = 0; i < n; ++i) {
v[i] = readInt(minv, maxv);
if (i+1 < n) readSpace();
}
return v;
}
auto readLongs(int n, long long minv, long long maxv) {
assert(n >= 0);
vector<long long> v(n);
for (int i = 0; i < n; ++i) {
v[i] = readLong(minv, maxv);
if (i+1 < n) readSpace();
}
return v;
}
void readSpace() {
assert((int) buffer.size() > pos);
assert(buffer[pos] == ' ');
pos++;
}
void readEoln() {
assert((int) buffer.size() > pos);
assert(buffer[pos] == '\n');
pos++;
}
void readEof() {
assert((int) buffer.size() == pos);
}
}inp;
#define NCR
#define PRIME M
int pw(int a,int p=M-2,int MOD=M){
int result = 1;
while (p > 0) {
if (p & 1)
result = a * result % MOD;
a = a * a %MOD;
p >>= 1;
}
return result;
}
int fact[200005],invfact[200005];
void init(){
int p=PRIME;
fact[0]=1;
int i;
for(i=1;i<200005;i++){
fact[i]=(i*fact[i-1])%p;
}
i--;
invfact[i]=pw(fact[i],p-2,p);
for(i--;i>=0;i--){
invfact[i]=(invfact[i+1]*(i+1))%p;
}
}
mint ncr(int n,int r){
if(r>n || r<0)return 0;
mint res = (((fact[n]*invfact[r])%PRIME)*invfact[n-r])%PRIME;
return res;
}
//x elements can't be at their pos,k can be anywhere
mint D(int x,int k){
int n = x+k;
mint tot = 0;
for(int i = 0;i <= x;i++){
if(i&1) tot -= ncr(x,i)*fact[n-i];
else tot += ncr(x,i)*fact[n-i];
}
return tot;
}
int smn = 0;
void solve()
{
int n = inp.readInt(1,200'000);
inp.readSpace();
smn += n;
int k = inp.readInt(0,n);
inp.readEoln();
vi a(n);
for(int i = 0;i < n;i++){
a[i] = inp.readInt(-1,n);
assert(a[i] != 0);
if(i == n-1)inp.readEoln();
else inp.readSpace();
}
int cnt = 0;
for(auto &x : a)if(x == -1)cnt++;
mint ans = fact[cnt];
set<int> s(be(a));
if(s.count(-1))s.erase(-1);
assert(s.size() == n-cnt);
mint p = 0;
if(n-k >= k){//last is not S
p = D(n-2*k,k)*ncr(n-k,k);
}
mint q = 0;
if(n-k >= k-1){//last is S
q = D(n-2*k+1,k-1)*ncr(n-k,k-1);
}
ans *= (p+q);
cout << ans.val() << "\n";
}
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#ifdef NCR
init();
#endif
#ifdef SIEVE
sieve();
#endif
// int t; cin >> t;
int t = inp.readInt(1,100'000);
inp.readEoln();
while(t--)
solve();
inp.readEof();
assert(smn <= 400'000);
return 0;
}