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: Wrong Answer)

#include <bits/stdc++.h>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
using ll = long long;
using P = pair<int,int>;
using Graph= vector<vector<int>>; 
#define rep(i,n) for (ll i=0; i < (n); ++i)
#define rep2(i,n,m) for(ll i=n;i<=m;i++)
#define rep3(i,n,m) for(ll i=n;i>=m;i--)
#define pb push_back
#define eb emplace_back
#define ppb pop_back
const ll INF=1e18 ;
inline void chmax(ll& a,ll b){a=max(a,b);}
inline void chmin(ll& a,ll b){a=min(a,b);}
// ミント
const int mod = 1e9+7 ;
//998244353;

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
//昆布
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
}   ;
 
mint modpow(ll a,ll b){
 if(b==0) return 1 ;
 mint c= modpow(a,b/2) ;
 if(b%2==1) return c*c*a ;
 else return c*c ;
}

int main(){
  ll k ;cin>> k ;
  string s ; cin>>s ;
  ll n=s.size() ;
  mint ans=1ll ;
  rep(i,26ll){
    ans= ans*26ll ;
  }
   cout<<ans<<endl ;
       return 0;
  }