NCOPRIMEN - Editorial

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

Setter: Satyam
Tester: Abhinav Sharma, Aryan
Editorialist: Lavish Gupta

DIFFICULTY:

Easy

PREREQUISITES:

Greatest Common Divisor

PROBLEM:

An array B of length N (N \ge 2) is said to be good if the following conditions hold:

  • For all 1 \leq i \leq N, 2 \leq B_i \leq 10^6
  • gcd(B_{i-1},B_i)\neq1 for all i (2 \leq i \leq N).

You have an array A of length N (2 \leq A_i \leq 10^5). You want to make the array A good.

To do so, you can change atmost \left \lceil{\frac {2⋅N} 3}\right \rceil elements of A.

Print the final array after changing A to a good array. If there are multiple possible final arrays, print any of them.

It is guaranteed that A can be made good after changing atmost \left \lceil{\frac {2⋅N} 3}\right \rceil elements of A.

CORNER CASE:

If you are getting Wrong Answer, the following test case might help you:

Corner Case

N = 6
A = \{ P_1, P_2, P_3, P_4, P_5, P_6 \} where P_i > 10^4, and each P_i is a prime number.

Make sure that your final array satisfies the condition A_i \leq 10^6, and you have not changed more than \left \lceil{\frac {2⋅N} 3}\right \rceil elements of A.

EXPLANATION:

This is one of the possible construction. Other solutions are also possible.

We are given that we can change atmost \left \lceil{\frac {2⋅N} 3}\right \rceil elements of A. Also note that the new elements should satisfy the following constraint: 2 \leq B_i \leq 10^6.
One useful way to interpret it is: out of every 3 elements, we can change 2 of them, such that the conditions are satisfied. Let us try if we can achieve the above conditions through this interpretation.

Let us consider the first 4 elements of the array - A_1, A_2, A_3 and A_4 and let each of the A_i be a prime number for the sake of this discussion.

Now, because gcd(A_1, A_2) = 1, we would like to change one of the A_1 or A_2. Let us greedily choose A_2 and try to change it to some optimal value. Let A_2 = \lambda \cdot A_1, for some \lambda.

Again, we may have gcd(A_2, A_3) = 1 depending on the value of \lambda, so we need to suitably change A_3. Let the new value of A_3 be A_3' .

Because we want to change only 2 values out of every 3 values, we don’t want to change our A_4. Hence, the following should also be satisfied: gcd(A_3', A_4) \neq 1. So, let us have A_3' = \lambda ' \cdot A_4.

So, the current array looks like \{A_1, \lambda \cdot A_1, \lambda ' \cdot A_4, A_4 , ...\}. The only condition that now needs to be satisfied is gcd(\lambda \cdot A_1, \lambda ' \cdot A_4) \neq 1.

Note that substituting \lambda = \lambda' = 2 will make the above property satisfied, as well as make sure that for each A_i, 2 \leq A_i \leq 10^6.

Hence, the final array would look like \{A_1, 2\cdot A_1, 2\cdot A_4, A_4, 2\cdot A_4, 2\cdot A_7, A_7, ...\}

TIME COMPLEXITY:

O(N) for each test case.

SOLUTION:

Setter's Solution

#include <bits/stdc++.h>   
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
#define ll long long  
const ll INF_MUL=1e13;
const ll INF_ADD=1e18;  
#define pb push_back               
#define mp make_pair        
#define nline "\n"                         
#define f first                                          
#define s second                                             
#define pll pair<ll,ll> 
#define all(x) x.begin(),x.end()   
#define vl vector<ll>         
#define vvl vector<vector<ll>>    
#define vvvl vector<vector<vector<ll>>>          
#ifndef ONLINE_JUDGE    
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#else
#define debug(x);  
#endif     
void _print(ll x){cerr<<x;}  
void _print(char x){cerr<<x;} 
void _print(string x){cerr<<x;}     
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); 
template<class T,class V> void _print(pair<T,V> p) {cerr<<"{"; _print(p.first);cerr<<","; _print(p.second);cerr<<"}";}
template<class T>void _print(vector<T> v) {cerr<<" [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T>void _print(set<T> v) {cerr<<" [ "; for (T i:v){_print(i); cerr<<" ";}cerr<<"]";}
template<class T>void _print(multiset<T> v) {cerr<< " [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T,class V>void _print(map<T, V> v) {cerr<<" [ "; for(auto i:v) {_print(i);cerr<<" ";} cerr<<"]";} 
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef tree<pair<ll,ll>, null_type, less<pair<ll,ll>>, rb_tree_tag, tree_order_statistics_node_update> ordered_pset;
//--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
const ll MOD=998244353;   
const ll MAX=500500;
ll binpow(ll a,ll b,ll MOD){
    ll ans=1;
    a%=MOD;
    while(b){
        if(b&1)
            ans=(ans*a)%MOD;
        b/=2;   
        a=(a*a)%MOD;
    }
    return ans;
}
ll inverse(ll a,ll MOD){
    return binpow(a,MOD-2,MOD);
}
void solve(){                
    ll n; cin>>n;
    vector<ll> a(n+5);   
    for(ll i=1;i<=n;i++){
        cin>>a[i];
    }
    a[n+1]=2;
    for(ll i=1;i<=n;i++){
        if((i%3)==1){
            continue;
        }
        if((i%3)==2){
            a[i]=a[i-1]*2;
        }
        else{
            a[i]=a[i+1]*2;
        }
    }
    for(ll i=1;i<=n;i++){
        cout<<a[i]<<" ";
    }
    cout<<nline;
    return;            
}                     
int main()                                                                      
{     
    ll test_cases=1;                   
    cin>>test_cases;
    while(test_cases--){   
        solve();     
    }
    cout<<fixed<<setprecision(10);
    cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n"; 
} 
First Tester's Solution
/* in the name of Anton */

/*
  Compete against Yourself.
  Author - Aryan (@aryanc403)
  Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
    #include <header.h>
#else
    #pragma GCC optimize ("Ofast")
    #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
    //#pragma GCC optimize ("-ffloat-store")
    #include<bits/stdc++.h>
    #define dbg(args...) 42;
#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
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

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
}

long long readInt(long long l, long long r, char endd) {
    long long x=0;
    int cnt=0;
    int fi=-1;
    bool is_neg=false;
    while(true) {
        char g=getchar();
        if(g=='-') {
            assert(fi==-1);
            is_neg=true;
            continue;
        }
        if('0'<=g&&g<='9') {
            x*=10;
            x+=g-'0';
            if(cnt==0) {
                fi=g-'0';
            }
            cnt++;
            assert(fi!=0 || cnt==1);
            assert(fi!=0 || is_neg==false);

            assert(!(cnt>19 || ( cnt==19 && fi>1) ));
        } else if(g==endd) {
            if(is_neg) {
                x=-x;
            }
            assert(l<=x&&x<=r);
            return x;
        } else {
            assert(false);
        }
    }
}
string readString(int l, int r, char endd) {
    string ret="";
    int cnt=0;
    while(true) {
        char g=getchar();
        assert(g!=-1);
        if(g==endd) {
            break;
        }
        cnt++;
        ret+=g;
    }
    assert(l<=cnt&&cnt<=r);
    return ret;
}
long long readIntSp(long long l, long long r) {
    return readInt(l,r,' ');
}
long long readIntLn(long long l, long long r) {
    return readInt(l,r,'\n');
}
string readStringLn(int l, int r) {
    return readString(l,r,'\n');
}
string readStringSp(int l, int r) {
    return readString(l,r,' ');
}

void readEOF(){
    assert(getchar()==EOF);
}

vi readVectorInt(int n,lli l,lli r){
    vi a(n);
    for(int i=0;i<n-1;++i)
        a[i]=readIntSp(l,r);
    a[n-1]=readIntLn(l,r);
    return a;
}

// #include<atcoder/dsu>
// vector<vi> readTree(const int n){
//     vector<vi> e(n);
//     atcoder::dsu d(n);
//     for(lli i=1;i<n;++i){
//         const lli u=readIntSp(1,n)-1;
//         const lli v=readIntLn(1,n)-1;
//         e[u].pb(v);
//         e[v].pb(u);
//         d.merge(u,v);
//     }
//     assert(d.size(0)==n);
//     return e;
// }

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
mt19937 rng(seed=chrono::steady_clock::now().time_since_epoch().count());
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;

    lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
    lli m;
    string s;
    vi a;
    //priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .

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);
T=readIntLn(1,1e5);
lli sumN = 2e5;
while(T--)
{

    n=readIntLn(2,min(100000LL,sumN));
    sumN-=n;
    a=readVectorInt(n,2,1e5);
    for(lli i=0;i<n;i+=3){
        if(i+1==n)
            a[i]=2*a[i-1];
        else
            a[i]=2*a[i+1];
    }
    for(lli i=2;i<n;i+=3)
        a[i]=2*a[i-1];
    for(auto x:a)
        cout<<x<<" ";
    cout<<endl;
    dbg(a);
    fo(i,n-1)
        assert(__gcd(a[i],a[i+1])!=1);
}   aryanc403();
    readEOF();
    return 0;
}

Second Tester's Solution
#include <bits/stdc++.h>
using namespace std;
 
 
/*
------------------------Input Checker----------------------------------
*/
 
long long readInt(long long l,long long r,char endd){
    long long x=0;
    int cnt=0;
    int fi=-1;
    bool is_neg=false;
    while(true){
        char g=getchar();
        if(g=='-'){
            assert(fi==-1);
            is_neg=true;
            continue;
        }
        if('0'<=g && g<='9'){
            x*=10;
            x+=g-'0';
            if(cnt==0){
                fi=g-'0';
            }
            cnt++;
            assert(fi!=0 || cnt==1);
            assert(fi!=0 || is_neg==false);
 
            assert(!(cnt>19 || ( cnt==19 && fi>1) ));
        } else if(g==endd){
            if(is_neg){
                x= -x;
            }
 
            if(!(l <= x && x <= r))
            {
                cerr << l << ' ' << r << ' ' << x << '\n';
                assert(1 == 0);
            }
 
            return x;
        } else {
            assert(false);
        }
    }
}
string readString(int l,int r,char endd){
    string ret="";
    int cnt=0;
    while(true){
        char g=getchar();
        assert(g!=-1);
        if(g==endd){
            break;
        }
        cnt++;
        ret+=g;
    }
    assert(l<=cnt && cnt<=r);
    return ret;
}
long long readIntSp(long long l,long long r){
    return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
    return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
    return readString(l,r,'\n');
}
string readStringSp(int l,int r){
    return readString(l,r,' ');
}
 
 
/*
------------------------Main code starts here----------------------------------
*/
 
const int MAX_T = 1e5;
const int MAX_N = 1e5;
const int MAX_SUM_LEN = 1e5;
 
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ff first
#define ss second
#define mp make_pair
#define ll long long
#define rep(i,n) for(int i=0;i<n;i++)
#define rev(i,n) for(int i=n;i>=0;i--)
#define rep_a(i,a,n) for(int i=a;i<n;i++)
 
int sum_len = 0;
int max_n = 0;
int yess = 0;
int nos = 0;
int total_ops = 0;

const ll mod = 998244353;

ll po(ll x, ll n ){ 
    ll ans=1;
     while(n>0){
        if(n&1) ans=(ans*x)%mod;
        x=(x*x)%mod;
        n/=2;
     }
     return ans;
}


void solve()
{   

    int n = readIntLn(2, 1e5);
    sum_len += n;
    max_n = max(max_n, n);
    
    int a[n+1];
    rep(i,n-1) a[i] = readIntSp(2, 1e5);
    a[n-1] = readIntLn(2, 1e5);
    a[n] = 1;

    rep_a(i,1,n){
        if(i%3==0) continue;
        else if(i%3==1) a[i] = 2*a[i-1];
        else a[i] = 2*a[i+1];
    }

    rep(i,n) cout<<a[i]<<" ";
    cout<<'\n';

}
 
signed main()
{

    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r" , stdin);
    freopen("output.txt", "w" , stdout);
    #endif
    fast;
    
    int t = 1;
    
    t = readIntLn(1,1e5);
        
    for(int i=1;i<=t;i++)
    {    
       solve();
    }
   
    assert(getchar() == -1);
    assert(sum_len<=2e5);
 
    cerr<<"SUCCESS\n";
    cerr<<"Tests : " << t << '\n';
    cerr<<"Sum of lengths : " << sum_len << '\n';
    cerr<<"Maximum length : " << max_n << '\n';
    // cerr<<"Total operations : " << total_ops << '\n';
    //cerr<<"Answered yes : " << yess << '\n';
    //cerr<<"Answered no : " << nos << '\n';
}

Editorialist's Solution
#include<bits/stdc++.h>
#define ll long long
using namespace std ;
const ll z = 998244353 ;


 
void solve()
{
    ll n;
    cin >> n ;
    ll arr[n] ;
    for(int i = 0 ; i < n ; i++)
        cin >> arr[i] ;

    for(int i = 0 ; i < n ; i++)
    {
        if(i%3 == 0)
            continue ;
        if(i%3 == 1)
            arr[i] = arr[i-1]*2 ;
        else
        {
            if(i+1 < n)
                arr[i] = arr[i+1]*2 ;
            else
                arr[i] = arr[i-1] ;
        } 
    }
    for(int i = 0 ; i < n ; i++)
        cout << arr[i] << ' ';
    cout << '\n' ;
    return ;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    #ifndef ONLINE_JUDGE
    freopen("inputf.txt" , "r" , stdin) ;
    freopen("outputf.txt" , "w" , stdout) ;
    freopen("error.txt" , "w" , stderr) ;
    #endif

    ll t;
    cin >> t ;
    while(t--)
    {
        solve() ;
    }

    return 0;
}
11 Likes

Nice Problem sir… failed but learnt <3

6 Likes

Good Problem Set… Couldn’t understand durning the contest after that learned it :smile:

2 Likes

I tried it using prime sieve but I failed :slight_smile:

1 Like

Can someone verify my logic about this case, it’s similar to the editorial but I got WA

  1. I would change every A[i], where i is odd to be A[i-1]*A[i+1], hence the gcd(A[i-1],A[i]) == A[i-1] and gcd(A[i],A[i+1]) == A[i+1]
  2. If length of A is even, i’ll change the last element in the array as the second last array
    If it is odd, i won’t change the last array

the end result for odd and even array is as following:
odd Array = A1, A1 x A2, A2
even Array = A1, A1 x A2, A2, A2

It is guaranteed to change at most 1/2+1 element which is less than 2/3N

Edit: I didn’t realize there is a constraint of A1 <= 10^6, it’s all cleared up. It’s just hard to understand that many notations at a glance lol

4 Likes

i did the same what’s wrong in this solution can anyone explain.

a[i] *a[i+1] should be less or equal than 10^6.
let suppose if we have 3 numbers 999998 , 99999, 10^5 for this case a[I] *a[i+1] >10^6

In your logic…you are not considering the edge cases, let consider A1 is some prime number ( A1 >= 10^4) and similarly A2 is some prime number (A2 >= 10^4), So in this case when you multiply both the number to replace it with middle element their product would be greater than 10^8, but in the question they have clearly mentioned that the resultant array elements can lie between 2 and 10^6 (2<=Bi<=10^6)

Let {A1,A2,A3,A4,A5,A6,A7} be the original given array, consider A1 is a prime number greater than 10^4 (A1>=10^4) and A3 is a prime number greater than 10^4 (A3 >= 10^4). Now going with your logic, you replace A2 with A1A3 (A1A3 >= 10^8), after multiplying both of them A2 becomes greater than 10^8 (A2>=10^8).

This was the observation i missed. I was trying about changing half of the indices and doing something with it. I upsolved it. Very beautiful problem and i mean it!!

I thought I had it all figured out but it was the corner cases :confused:
Learned from this one.

1
7
30103 30203 30403 30703 30803 31013 31513

I read the corner case hint and believe me i knew that before
but i don’t know how to solve this corner case even manually
i don’t want to read the answer editorial now, can you help me with this corner case??

edit : nvm i looked at the editorial :man_facepalming:

It was a beautiful problem, enjoyed solving it!


// Non coprime neighbours

#include<bits/stdc++.h>
#define int long long int
#define vi vector<int>
#define vii vector<vi>
#define pii pair<int, int>
#define min_pq priority_queue<int, vector<int>, greater<int>>
#define take_input freopen("input.txt", "r", stdin)
#define give_output freopen("output.txt", "w", stdout)

using namespace std;

vi smallestprime(int n=1e6) {
	vi isp(n+1);
	for(int i=0; i<=n; i++) isp[i] = i;
	for(int i=2; i*i<=n; i++){
		for(int j=i*i; j<=n; j+=i) {
			if(isp[j] == j) {
				isp[j] = i;
			}
		}
	}
	return isp;
}

int32_t main() {
	// take_input;
	// give_output;
	vi isp = smallestprime();
	int t; cin >> t;
	while(t--){
		int n; cin >> n;
		vi arr(n);
		for(int &i:arr) cin >> i;
		int limit = ((2*n)%3 == 0? 2*n/3:2*n/3+1);
		int cnt=0;
		for(int i=1; i<n; i++) {
			if(__gcd(arr[i-1], arr[i]) == 1){
				if(i!=n-1){
					arr[i] = isp[arr[i-1]]*isp[arr[i+1]];
				}else{
					arr[i] = arr[i-1];
				}
				cnt++;
			}
		}
		if(cnt <= limit) {
			for(int i:arr) cout << i << " ";
			cout << endl;
		}
	}	
}

I got the solution but here I am getting WA for all the test set. except one. Can somebody explain this. I first tried arr[i] = arr[i-1]*arr[i+1]/__gcd(arr[i-1], arr[i+1]) but because of the limit changed it to this but no difference at all

Hi @chetan06 lets you have 2 prime number one at arr[i-1] and one at arr[i+1] and lets say these prime numbers are of the order of 10^5 which is the range given in the question then their product will be of order 10^10 and gcd of two prime numbers is always 1 so this arr[i-1]*arr[i+1]/__gcd(arr[i-1], arr[i+1]) will be order of 10^10 which is greater than 10^6 hence WA

Ohh that was silly :sweat_smile: By the way, thanks

1 Like
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define endl "\n"
#define MOD 1e9 + 7
#define int int64_t
using namespace std;
int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    int Tc;
    cin >> Tc;
    while (Tc--)
    {
        int n;
        cin >> n;
        vector<int> vec(n);
        for (auto &&i : vec)
            cin >> i;

        for (int i = 1; i < n; i += 2)
        {
            int x = vec[i - 1];
            int y = vec[i + 1];
            int gcd = __gcd(x, y);

            if (i == n - 1)
            {
                vec[i] = x * vec[i];
                vec[i] = __gcd(vec[i - 1], vec[i]);
                continue;
            }

            if (x % 2 == 0 and y % 2 == 0 and gcd > 1)
            {
                vec[i] = gcd;
            }
            else if ((x + y) & 1)
            {
                if (x & 1)
                {
                    vec[i] = 2 * x;
                }
                else
                {
                    vec[i] = 2 * y;
                }
            }
            else
            {
                int z = x * y;
                vec[i] = __gcd(z, x) * __gcd(z, y);
            }
        }

        for (auto &&i : vec)
        {
            cout << i << " ";
        }
        cout << endl;
    }

    return 0;
}

can someone tell me what’s wrong in this code :frowning:

Can anyone tell me why i am getting WA with this approach

void solve() {
	int n;
	cin >> n;
	vector<int> v(n + 2);
	for (int i = 1; i < n + 1; i++) {
		cin >> v[i];
	}
	int i = 1;
	while (i <= n) {
		if (i >= n - 1) {
			v[n - 1] = 2 * v[n];
			break;
		}
		if (v[i + 1] & 1) {
			v[i] = v[i + 1] * 2;
			v[i + 2] = v[i + 1] * 2;
		}
		else {
			if (v[i] & 1)
				v[i] = 2 * v[i + 1];
			if (v[i + 2] & 1)
				v[i + 2] = 2 * v[i + 1];
		}
		i += 3;
	}
	for (int i = 1; i <= n; i++) {
		cout << v[i] << " ";
	} cout << endl;
}

Input

1
4
3 5 2 3

Output
10 5 6 3

Still WA after updating:

if (i >= n - 1) {
			v[n] = 2 * v[n - 2];
			v[n - 1] = 2 * v[n - 2];
			break;
		}

Input

1
2
2 3

Output:
0 0

1 Like