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<bits/stdc++.h>
using namespace std;
const int Max_N=1e6; 
const long long Mod=(1e9)+7;
int n,m;
char s[Max_N+10];
long long ans=0,p[Max_N*2+10]={},inv[Max_N*2+10]={};
inline long long Mul(int x,int y){
	long long a=1,b=x;
	for(;y;b=b*b%Mod,y>>=1) if(y&1) a=a*b%Mod;
	return a; 
}
inline long long C(int n,int m){
	return p[n]*inv[m]%Mod*inv[n-m]%Mod;
}
int main(){
	scanf("%d",&n); scanf("%s",s); m=strlen(s);
	p[0]=1; for(int i=1;i<=n+m;i++) p[i]=p[i-1]*i%Mod;
	inv[0]=inv[1]=1; for(int i=2;i<=n+m;i++) inv[i]=inv[Mod%i]*(Mod-Mod/i)%Mod;
	for(int i=1;i<=n+m;i++) inv[i]=inv[i-1]*inv[i]%Mod;
	for(int i=0;i<=n;i++) ans+=C(n+m-i-1,m-1)*Mul(26,i)%Mod*Mul(25,n-i)%Mod,ans%=Mod;
	printf("%lld\n",ans);
	return 0;
}