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++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

const int mod = 1000000007;
// const int mod = 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; }

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];
  }
};

int main() {
  int k, n;
  string s;
  cin >> k >> s;
  n = s.size();
  combination c(n + k);
  mint ans = 0;
  for (int i = 0; i <= k; i++) {
    mint now = mint(26).pow(k - i);
    now *= mint(25).pow(i);
    now *= c(i + n - 1, n - 1);
    ans += now;
  }
  cout << ans << endl;
  return 0;
}