How many strings can be obtained by applying the following operation on a string S exactly K times: "choose one lowercase English letter and insert it somewhere"?

The answer can be enormous, so print it modulo (10^9+7).

Constraints

* K is an integer between 1 and 10^6 (inclusive).
* S is a string of length between 1 and 10^6 (inclusive) consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:


K
S


Output

Print the number of strings satisfying the condition, modulo (10^9+7).

Examples

Input

5
oof


Output

575111451


Input

37564
whydidyoudesertme


Output

318008117
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

const int N=2000005;
const int mod=1000000007;

int n,m,ans,sum;
int fac[N],inv[N];
char s[N];

int quickpow(int x,int y){
	int s=1;
	for (;y;y>>=1,x=1ll*x*x%mod)
		if (y&1) s=1ll*s*x%mod;
	return s;
}

void init(int l){
	fac[0]=inv[0]=1;
	for (int i=1;i<=l;i++) fac[i]=1ll*fac[i-1]*i%mod;
	inv[l]=quickpow(fac[l],mod-2);
	for (int i=l-1;i;i--) inv[i]=1ll*inv[i+1]*(i+1)%mod;
}

int C(int n,int m){
	if (n<m) return 0;
	return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}

int main(){
	scanf("%d",&n);
	scanf("%s",s);
	m=strlen(s);
	init(n+m);
	ans=0;
	sum=1;
	for (int i=1;i<=n+m&&n+m-i-m+1>=0;i++){
		(ans+=1ll*sum*quickpow(25,n+m-i-(m-1))%mod*C(n+m-i,m-1)%mod)%=mod;
		sum=1ll*sum*26%mod;
	}
	printf("%d\n",ans);
	return 0;
}