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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <math.h>
#include <numeric>
#include <complex>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <functional>
#include <assert.h>
using ll=long long;
#define int ll
#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define eb emplace_back
#define all(x) x.begin(),x.end()
#define SZ(x) ((int)x.size())
#define fi first
#define se second
const int INF = 1e15;
const int MOD = 1e9+7;
using namespace std;

template<class T> using vc=vector<T>;
template<class T> using vvc=vc<vc<T>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class T>
inline bool chmax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<class T>
inline bool chmin(T &a,T b){if(a>b){a=b;return true;}return false;}

const int MX = 2000100 + 5;
int fac[MX+5],inv[MX+5];
 
int power(int x,int y,int p){
	int res=1;
	x=x%p;
	while(y){
		if(y&1)res=(res*x)%p;
		y=y>>1;
		x=(x*x)%p;
	}
	return res;
}
void init(){
  int n=MX+4;
	fac[0] = 1;
  for(int i=0;i<n;++i)fac[i+1]=fac[i]*(i+1)%MOD;
  inv[n]=power(fac[n],MOD-2,MOD);
  for(int i=n-1;i>=0;--i)inv[i]=inv[i+1]*(i+1)%MOD;
}
int comb(int n,int r){return (fac[n]*inv[r])%MOD*inv[n-r]%MOD;}

signed main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout<<fixed<<setprecision(20);
    init();

    int K; cin >> K;
    string S; cin >> S;
    int T=K+SZ(S);
    int ans=0;
    rng(i,SZ(S),K+SZ(S)+1) {
        ans+=(comb(T,i)*power(25,T-i,MOD))%MOD;
        ans%=MOD;
    }
    cout << ans << endl;
}