Problem:
Score: 600 points

Problem Statement

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).

Sample Input 1

5
oof

Sample Output 1

575111451

For example, we can obtain proofend, moonwolf, and onionpuf, while we cannot obtain oofsix, oofelevennn, voxafolt, or fooooooo.

Sample Input 2

37564
whydidyoudesertme

Sample Output 2

318008117

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(10);}} __init;
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
constexpr int MOD=1000000007;
//constexpr int MOD=998244353;
#define INF (1<<30)
#define LINF (lint)(1LL<<56)
#define endl "\n"
#define rep(i,n) for(lint (i)=0;(i)<(n);(i)++)
#define reprev(i,n) for(lint (i)=(n-1);(i)>=0;(i)--)
#define Flag(x) (1<<(x))
#define Flagcount(x) __builtin_popcount(x)
#define pint pair<int,int>
#define pdouble pair<double,double>
#define plint pair<lint,lint>
#define fi first
#define se second
typedef long long lint;
int dx[8]={1,1,0,-1,-1,-1,0,1};
int dy[8]={0,1,1,1,0,-1,-1,-1};
const int MAX_N=2e5+5;

lint fac[2100000],finv[2100000],inv[2100000];
void COMinit(){
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for(int i=2;i<2100000;i++){
    fac[i]=fac[i-1]*i%MOD;
    inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
    finv[i]=finv[i-1]*inv[i]%MOD;
  }
}
lint COM(lint n,lint k){
  if (n<k) return 0;
  if (n<0 || k<0) return 0;
  return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}
lint facdiv(lint n,lint k){
  if (n<k) return 0;
  if (n<0 || k<0) return 0;
  return (fac[n]*finv[k])%MOD;
}

lint powmod(lint a,lint b){
  return b?powmod(a*a%MOD,b/2)*(b%2?a:1)%MOD:1;
}

int main(void){
    COMinit();
    lint K;
    string S;
    cin >> K >> S;
    lint N=S.length();
    lint ans=0;
    for(int i=0;i<=K;i++){
        lint add=(powmod(26,i)*powmod(25,K-i))%MOD;
        add=add*COM(N+K-1-i,N-1);
        ans=(ans+add%MOD)%MOD;
    }
    cout << ans << endl;
}