stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include<stdio.h>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

for(int i=1;i<=N;i++)

{a[i]+=K; a[i]-=Q; }

for(int i=1;i<=Q;i++)

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

	while(cin >> n >> k >> q)

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

		for(long long int i = 1; i <= q; i++)

		{

			scanf("%lld", &m);

			A[m]++;

		}

		for(long long int i = 1; i <= n; i++)

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



const int MAXN = 100005;



int N, K, Q, ai, A[MAXN];

int main(){

	scanf("%d %d %d", &N, &K, &Q);

	for(int i = 0; i < Q; i++){

		scanf("%d", &ai);

		A[ai - 1]++;

	}

	for(int i = 0; i < N; i++){

		if(K - Q + A[i] > 0)printf("Yes\n");

		else printf("No\n");

	}

	

	return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 3;

int n, k, q, freq[N];

int main() {

#ifndef ONLINE_JUDGE

//	freopen("input.in", "r", stdin);

#endif

	scanf("%d %d %d", &n, &k, &q);

	for (int i = 0; i< q; ++i) {

		int a;

		scanf("%d", &a);

		++freq[a];

	}

	for (int i = 1; i <=n; ++i) {

		if (k - (q - freq[i]) > 0) {

			puts("Yes");

		}else {

			puts("No");

		}

	}

	return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define ll long long

using namespace std;

ll n , k ,q ,a[100001] ,x ;

int main()

{

    cin >> n >> k >> q ;

    for (int  i = 1 ;i <= n ;i++)

    {

        a[i] += k;

        a[i] -= q ;

    }

    while (q--)

    {

     cin >> x ;

        a[x]+=1;

    }

    for (int  i = 1 ;i <= n ; i++)

    {

        if (a[i]  > 0 )cout <<"Yes"<<endl;

        else cout <<"No"<<endl;

    }

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<iostream>

using namespace std;

const int N=100010;

int n,k,q,pl[N];

int main()

{

	scanf("%d%d%d",&n,&k,&q);

	for(int i=1;i<=n;i++) pl[i]=k;

	for(int i=1;i<=q;i++)

	{

		int x;

		scanf("%d",&x);

		pl[x]++;

	}

	for(int i=1;i<=n;i++)

	{

		if(pl[i]-q>0) printf("Yes\n");

		else printf("No\n");

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <stdio.h>

#include <iostream>

using namespace std;



int N,K,Q,a,nilai;

int main (){

int nilai[100001];

cin >> N >> K >>Q;

	for (int i=1; i<=N; i++){

		nilai[i]=K-Q;

	}

	for (int i=1; i<=Q; i++){

		cin>>a;

		nilai[a]++;

	}	

	for (int i=1; i<=N; i++){

		if (nilai[i]<=0){

			cout << "No"<< endl;

		}else {

			cout <<"Yes"<< endl;

		}

	}

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;



long long n,k,q,bil[100001],a;



int main(){

    cin>>n>>k>>q;

    for(int h=1;h<=n;h++){

    bil[h]=k-q;

    }

    for(int i=1;i<=q;i++){

        cin>>a;

        bil[a]++;

    }

    for(int i=1;i<=n;i++){

        if(bil[i]<=0){

            cout<<"No"<<endl;

        }else{

            cout<<"Yes"<<endl;

        }

    }

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <algorithm>

#include <cstring>

#include <cmath>

#include <cstdio>

#include <vector>

#include <queue>



using namespace std;



typedef long long ll;

const int maxn=1e5+7;



int a[maxn];



int main()

{

//    freopen("input.txt","r",stdin);

    

    int n,k,q;

    cin>>n>>k>>q;

    for(int i=0;i<n;i++) a[i]=k-q;

    for(int i=0;i<q;i++)

    {

    	int x;

    	cin>>x;

    	a[x-1]++;

	}

	for(int i=0;i<n;i++) 

	{

		if(a[i]>0) cout<<"Yes"<<endl;

		else cout<<"No"<<endl;

	}

    

    return 0;

}

//想着稍微改一改思路来做这道题也不错 

//考虑一下我为什么会把G题的思路想起来 


























// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<stack>

#include<cstring>

#include<cmath>

#include<queue>

#include<set>

using namespace std;

typedef long long ll;

ll a[1111111];

ll b[1111111];

ll c[1111111];



int main()

{

	//ios::sync_with_stdio(false);

	ll n,k,q,x;

	cin>>n>>k>>q;

	for(ll i=0;i<n;i++)

		a[i]=k;

	set<ll> st; 

	for(ll i=0;i<q;i++)

	{

		scanf("%lld",&x);

		a[x-1]++;

	}

	



	for(ll j=0;j<n;j++)

		if(a[j]-q>0)

			cout<<"Yes"<<endl;

		else

			cout<<"No"<<endl;

	return 0;

}




// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

    for(int i = 1 ; i <= n ; i ++ )

        a[i] = k;

    for(int i = 1 ; i <= q ; i ++ )

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        a[i] -= q;

    }

    for(int i = 1 ; i <= n ; i ++ )

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>

#include <cstdio>

#include <cstring>

#include <ctime>

#include <string>

#include <vector>

#include <algorithm>

#include <cmath>

#include <sstream>

#include <utility>

using namespace std;

typedef long long ll;

typedef vector<int> vi;

typedef vector<vi> vvi;

typedef pair<int,int> pii;

#define all(c) (c).begin(), (c).end()

#define loop(i,a,b) for(ll i=a; i<ll(b); i++)

#define rep(i,b) loop(i,0,b)

#define each(e,c) for(auto&e:c)

#define pb push_back

#define eb emplace_back

#define mp make_pair

#define mt make_tuple

#define lb lower_bound

#define ub upper_bound

#ifdef DEBUG

#define dump(...) (cerr<<#__VA_ARGS__<<" = "<<(DUMP(),__VA_ARGS__).str()<<" ["<<__LINE__<<"]"<<endl)

struct DUMP:ostringstream{template<class T>DUMP &operator,(const T&t){if(this->tellp())*this<<", ";*this<<t;return *this;}};

#else

#define dump(...)

#endif

template<class T> ostream& operator<<(ostream& os, vector<T> const& v){

    rep(i,v.size()) os << v[i] << (i+1==v.size()?"":" ");

    return os;

}



int main(){

    string s;

    while(cin>>s){

    int n = s.size();

    vi l(n);

    l[0] = 1;

    rep(i,n-1){

        if(s[i+1]==s[i]) l[i+1]=l[i]+1;

        else l[i+1] = 1;

    }

    int ans = 0;

    dump(s);

    dump(l);

    loop(i,1,n-1){

        if(s[i]=='O' && s[i+1]!='O' && i+l[i]<n && i-l[i] >= 0 && l[i+l[i]]>=l[i] && l[i-l[i]]>=l[i] && s[i+l[i]]=='I' && s[i-l[i]]=='J'){

            ans = max(ans,l[i]);

        }

    }

    cout << ans << endl;

}

}

// Optimized solution to the same problem (faster):
#include <cstring>

#include <cstdio>

#include <algorithm>

using namespace std;

#define loop(i,a,b) for(int i=a; i<int(b); i++)

#define rep(i,b) loop(i,0,b)



char s[1000100];

int l[1000100];

int main(){

    scanf("%s",s);

    int n = strlen(s);

    l[0] = 1;

    rep(i,n-1){

        if(s[i+1]==s[i]) l[i+1]=l[i]+1;

        else l[i+1] = 1;

    }

    int ans = 0;

    loop(i,1,n-1){

        if(s[i]=='O' && s[i+1]!='O' && i+l[i]<n && i-l[i] >= 0 && l[i+l[i]]>=l[i] && l[i-l[i]]>=l[i] && s[i+l[i]]=='I' && s[i-l[i]]=='J'){

            ans = max(ans,l[i]);

        }

    }

    printf("%d\n",ans);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <cstdio>

#include <cstring>

#include <string>

#include <algorithm>

using namespace std;

char x[1000005];

int a=0,b=0,c=0,d=0,e,f,i,j,k,maxi=0;

int main(){

	fill(x,x+1000005,'A');

	scanf("%s",x);

int leng=strlen(x);

while(1){

if(a==leng){break;}

if(x[a]!='J'){

a++;

}else{

  while(1){

    if(x[a]=='J'){

      b++;

      a++;

    }else{

       break;

}}

while(1){

if(x[a]=='O'){

c++;

a++;

}else{

break;

}}

while(1){

if(x[a]=='I'){

d++;

a++;

}else{

break;

}}

if(b>=c && c<=d){

maxi=max(maxi,c);

}

b=0;

c=0;

d=0;

}

}

printf("%d\n",maxi);scanf("%d",&a);

return 0;

}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

#include <vector>

#include <cstring>

using namespace std;

char str[1000005];

int main()

{

  scanf("%s",&str);

  int leng=strlen(str);

  int ans=0,i=0;

 for(;;)

    {

      if(str[i]=='J')

	{

	  int a=0,b=0,c=0;

	  while(str[i]=='J' && i<leng)

	    {

	      i++; a++;

	    }

	  while(str[i]=='O' && i<leng)

	    {

	      i++;

	      b++;

	    }



	  while(str[i]=='I' && i<leng)

	    {

	      i++;

	      c++;

	    }

	  if(b<=a && b<=c)

	    {

	      ans=max(ans,b);

	    }

		if(i==leng) break;

	}else{

	    i++;

            if(i==leng) break;

	}



    }



  printf("%d\n",ans);

  return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
//Bokan ga bokka--nn!!

//Daily Lunch Special Tanoshii !!

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <string>

#include <algorithm>

#include <vector>

#include <queue>

#include <stack>

#include <functional>

#include <iostream>

#include <map>

#include <set>

using namespace std;

typedef pair<int,int> P;

typedef pair<int,P> P1;

typedef pair<P,P> P2;

typedef long long ll;

#define pu push

#define pb push_back

#define mp make_pair

#define eps 1e-7

#define INF 2000000000

#define s(x) scanf("%d",&x)

#define rep(i,x) for(int i=0;i<x;i++)

int main()

{

	string s;

	cin >> s;

	int id=0;

	int ret=0;

	while(id<s.size())

	{

		if(s[id]!='J')

		{

			id++;

			continue;

		}

		int j=0,o=0,i=0;

		while(id<s.size() && s[id]=='J') { j++; id++;}

		while(id<s.size() && s[id]=='O') { o++; id++;}

		while(id<s.size() && s[id]=='I') { i++; id++;}

		if(j<o || o>i) continue;

		ret=max(ret,o);

	}

	printf("%d\n",ret);

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <string.h>

char s[1000005];

int min(int a,int b){ return a<b?a:b; }

int max(int a,int b){ return a>b?a:b; }

int main(){

	scanf("%s",&s); int n=strlen(s),beg=0,ret=0;

	while(beg<n)

	{

		if(s[beg]!='J') { beg++; continue; }

		int j=0,o=0,i=0;

		while(beg<n && s[beg]=='J'){ j++; beg++;}

		while(beg<n && s[beg]=='O'){ o++; beg++;}

		while(beg<n && s[beg]=='I'){ i++; beg++;}

		if(o==min(j,min(o,i)))

		{

			ret=max(ret,o);

		}

	}

	printf("%d\n",ret);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <string>

#include <vector>

#include <iostream>

#include <algorithm>



using namespace std;



int main()

{

    string S; cin >> S;

    

    vector<char> v1;

    

    vector<int> v2; int c = 1;

    

    for(int i = 1; i < S.size(); i++)

    {

        if(S[i - 1] != S[i])

        {

            v1.push_back(S[i - 1]);

            v2.push_back(c);

            

            c = 0;

        }

        

        c++;

    }

    

    v1.push_back(S[S.size() - 1]);

    v2.push_back(c);

    

    int ret = 0;

    

    for(int i = 1; i < v1.size() -1; i++)

    {

        if(v1[i - 1] == 'J' && v1[i] == 'O' && v1[i + 1] == 'I' && v2[i - 1] >= v2[i] && v2[i] <= v2[i + 1])

        {

            ret = max(ret, v2[i]);

        }

    }

    

    printf("%d\n", ret);

    

    return 0;

}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>

char s[1000009], t[1000009]; int c[1000009];

int main() {

	scanf("%s", s);

	int n = strlen(s);

	int p = 0, z = 0;

	for (int i = 1; s[i - 1]; i++) {

		if (s[i] != s[i - 1]) {

			t[z] = s[i - 1];

			c[z] = i - p;

			p = i; z++;

		}

	}

	int ret = 0;

	for (int i = 1; i < z - 1; i++) {

		if (t[i - 1] != 'J' || t[i] != 'O' || t[i + 1] != 'I') continue;

		if (c[i - 1] < c[i] || c[i] > c[i + 1]) continue;

		if (ret < c[i]) ret = c[i];

	}

	printf("%d\n", ret);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

 

typedef long double ld;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> i_i;

typedef pair<ll, int> ll_i;

typedef pair<int, ll> i_ll;

typedef pair<double, int> d_i;

typedef pair<ll, ll> ll_ll;

typedef pair<double, double> d_d;

struct edge { int u, v; ll w; };

 

ll MOD = 1000000007;

ll _MOD = 1000000009;

double EPS = 1e-10;

 

int main() {

	int N; cin >> N;

	string s; cin >> s;

	int M = s.length();

	vector<int> dp(N + 1);

	dp[0] = 1;

	for (int t = 0; t < N; t++) {

		vector<int> _dp(N + 1);

		for (int x = 0; x <= t; x++) {

			_dp[x + 1] = (_dp[x + 1] + dp[x]) % MOD;

			if (x) _dp[x - 1] = (_dp[x - 1] + dp[x] * 2LL) % MOD;

			else _dp[x] = (_dp[x] + dp[x]) % MOD;

		}

/*		cout << "at " << t+1 << " phase:\n";

		cout << "dp: ";

		for(int j = 0 ; j <= N ; j ++) cout << dp[j] << " ";cout << endl;

		cout << "_dp: ";

		for(int j = 0 ; j <= N ; j ++) cout << _dp[j] << " ";cout << endl;

*/		dp = _dp;

	}

	cout << dp[M] << endl;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring>

#include<algorithm>

#define MAXN 5000

#define MO 1000000007

using namespace std;

char str[MAXN+5];

int N;

int dp[MAXN+5][MAXN+5];//dp[i][j]:走到第j位花了i步

int main()

{

	scanf("%d",&N);

	scanf("%s",str+1);

	int len=strlen(str+1);

//	int len=N/2;

	dp[0][0]=1;

	for(int tmn=1;tmn<=N;tmn++)

	{

		for(int pos=0;pos<=N;pos++)

		{

			if(pos!=N)

				dp[tmn][pos]=(1LL*dp[tmn][pos]+2LL*dp[tmn-1][pos+1]%MO)%MO;

			if(pos!=0)

				dp[tmn][pos]=(1LL*dp[tmn][pos]+1LL*dp[tmn-1][pos-1])%MO;

			if(pos==0)

				dp[tmn][pos]=(1LL*dp[tmn][pos]+1LL*dp[tmn-1][pos])%MO;

		}

	}

	printf("%d\n",dp[N][len]);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define LL long long

#define Mod 1000000007

using namespace std;

int n,m;

int f[2][5005];

int main(){

	char*str=new char[5005];

	scanf("%d%s",&n,str);

	m=strlen(str);

	delete [] str;

	f[0][0]=1;

	for(int i=1;i<=n;++i){

		int I=i&1;

		f[I][0]=(f[I^1][0]+f[I^1][1])%Mod;

		for(int j=1;j<=n;++j)

			f[I][j]=((LL)f[I^1][j-1]+f[I^1][j-1]+f[I^1][j+1])%Mod;

	}

	int Ans=f[n&1][m];

	for(int i=1;i<=m;++i) Ans=(LL)Ans*500000004%Mod;

	printf("%d",Ans);

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring>

#include<algorithm>

#define MAXN 5000

#define MO 1000000007

using namespace std;

char str[MAXN+5];

int N;

int dp[MAXN+5][MAXN+5];//dp[i][j]:走到第i位花了j步

int main()

{

	scanf("%d",&N);

	scanf("%s",str+1);

	int len=strlen(str+1);

//	int len=N/2;

	dp[0][0]=1;

	for(int tmn=1;tmn<=N;tmn++)

	{

		for(int pos=0;pos<=N;pos++)

		{

			if(pos!=N)

				dp[tmn][pos]=(1LL*dp[tmn][pos]+2LL*dp[tmn-1][pos+1]%MO)%MO;

			if(pos!=0)

				dp[tmn][pos]=(1LL*dp[tmn][pos]+1LL*dp[tmn-1][pos-1])%MO;

			if(pos==0)

				dp[tmn][pos]=(1LL*dp[tmn][pos]+1LL*dp[tmn-1][pos])%MO;

		}

	}

	printf("%d\n",dp[N][len]);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



#define int long long



using namespace std;



const int N = 5005;

const int mod = 1e9 + 7;



void add (int &x, int y) {

    x += y;

    x %= mod;

}



int dp[N], f[N];



signed main(){

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);



    int n;

    string s;

    cin >> n >> s;



    dp[0] = 1;

    for (int i = 0; i < n; i++) {

        memset(f, 0, sizeof(f));

        for (int j = 0; j <= i; j++) {

            add(f[j + 1], dp[j]);

            if (j) add(f[j - 1], 2 * dp[j]);

            else add(f[j], dp[j]);

        }



        for (int j = 0; j < N; j++) dp[j] = f[j];

    }



    cout << dp[s.length()];

    return 0;

}


// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring>

const long long mod=1e9+7;

long long matmod(long long val,long long k){

	long long ret=1ll;

	for(;k;k>>=1){

		if(k&1) ret=(val*ret)%mod;

		val=(val*val)%mod;

	}

	return ret;

}

void add(long long &des,long long sor){des+=sor;des%=mod;return ;}

char s[5005];

int n;

long long f[2][5005];

int main(){

	//freopen("keyboard.in","r",stdin);

	//freopen("keyboard.out","w",stdout);

	scanf("%d",&n);

	scanf("%s",s);

	int len=strlen(s);

	if(len>n){puts("0");return 0;}

	long long mul=matmod(2ll,len);

	f[0][0]=1ll;

	long long ans=f[0][len];

	for(int i=1;i<=n;i++){

		f[i&1][0]=(f[(i-1)&1][0]+f[(i-1)&1][1])%mod;

		for(int j=1;j<=i;j++){

			f[i&1][j]=0; 

			add(f[i&1][j],f[(i-1)&1][j-1]*2ll);

			add(f[i&1][j],f[(i-1)&1][j+1]);

		}

		add(ans,f[i&1][len]);

	}

	long long tmp=matmod(mul,mod-2ll);

	long long newans=f[n&1][len]*tmp%mod;

	printf("%lld\n",newans);

	return 0;

}

// Fastest known solution for this problem: