text: #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>;
  
  // 異なる操作が結局同じ結果になるような重複を取り除く、という方針では頓挫することが多い
  
  /*
      参考リンク
      ABC 171 F - Strivore
        https://atcoder.jp/contests/abc171/tasks/abc171_f
  */
  
  const int mod = 1000000007;
  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; }
  // combination mod prime
  // https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
  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;
  }