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)

#include <bits/stdc++.h>
#define REP(i,n) for(int i=0;i<n;i++)
#define REPP(i,n) for(int i=1;i<=n;i++)
#define ALL(obj) (obj).begin(), (obj).end()
long long INF = (long long)1E17;
#define i_7 (long long)(1E9+7)
//#define i_7 998'244'353
long mod(long a){
    long long c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}

//typedef long long ll;
long long po(long a, long b){
    if(b==0){
        return 1;
    }
    
    long long z = po(a,b/2);
    z = mod(z*z);
    if(b%2!=0){
        z = mod(a*z);
    }
    return z;
}

//配列のSIZEは問題ごとに定義する必要がある。
const int SIZE = 2000010;
long long inv[SIZE+1];//各iの逆元を格納する配列。
long long kai[SIZE+1];//i!のmodを格納する配列。
long long invkai[SIZE+1];//各i!の逆元を格納する配列。
const int MOD=i_7;
void invinit(){//上の配列を初期化する関数。
    inv[1]=1;
    for(int i=2;i<=SIZE;i++){
        inv[i] = MOD - ((MOD/i)*inv[MOD%i])%MOD;
    }
    
    kai[0]=invkai[0]=1;
    for(int i=1;i<=SIZE;i++){
        kai[i]=(kai[i-1]*i)%MOD;
        invkai[i]=(invkai[i-1]*inv[i])%MOD;
    }
}

long long comb(long long a, long long b){
    if(b<0 || a<b){
        return 0;
    }
    return mod(kai[a]*mod(invkai[b]*invkai[a-b]));
}

int main(){
  invinit();
  using namespace std;
  long long n, k;
  cin>>k;
  string s;
  cin>>s;
  n = s.size();
  long long ans = 0;
  for(long long i = 0; i <= k; i++){
    long long c = comb(n + k - i - 1, n - 1);
    long long temp = mod(po(25, k-i) * po(26, i));
    ans += mod(c * temp);
    ans = mod(ans);
  }
  cout << ans << endl;
  return 0;
}