Problem p02632 (C++)

Buggy submission (status: Runtime Error):

// Problem : F - Strivore
// Contest : AtCoder - AtCoder Beginner Contest 171
// URL : https://atcoder.jp/contests/abc171/tasks/abc171_f
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

/*
We want to count the number of strings T of length |S|+K such that the given
string S is a subsequence of T. Let's fix T and try to build an algorithm to
check if S is a subsequence. Our algorithm if successful will return
a1,a2,...,a|S| such that Si=Tai. Let's build a in a way such that ai is minimum
possible for all i. Note that for each i<|S|, characters Tai+1,...,Tai+1−1 can't
be equal to Tai+1 (otherwise we can minimize ai+1), so for those characters we
have limited choices (only 25). Characters after a|S| can be anything (26
choices).

So, let's loop over how many characters will be after a|S|; let that number be
x. We have 26x choices for that suffix. For each of the other K−x characters, we
only have 25 choices, so we have 25K−x choices for them in total. Still, we have
to decide how to distribute these K−x characters. We have |S| slots for them (
before the first character of S, between the first and the second character,
...). Using stars and bars, we can distribute them in (|S|−1−k−x|S|−1) ways. So,
we will loop over x and add to the answer 25K−x∗26x∗(|S|−1−k−x|S|−1).
*/

/* ID: vqt
LANG: C++14
PROB: namenum
*/

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>
#include <unistd.h>
using namespace std;

// for PDBS use only:
//  #include <ext/pb_ds/assoc_container.hpp>
//  #include <ext/pb_ds/tree_policy.hpp>
//  using namespace __gnu_pbds;
//  template <typename T> using o_set = tree<T, null_type, greater<T>,
//  rb_tree_tag, tree_order_statistics_node_update>;
typedef long long ll;
// #define ordered_set tree<pair<ll,ll>, null_type,less<pair<ll,ll>>,
// rb_tree_tag,tree_order_statistics_node_update>

// defines
#define int long long
// #define double long double
#define vec vector
#define vi vector<int>
#define vvi vector<vector<int>>
#define ii pair<int, int>
#define F first
#define S second
#define MP make_pair
#define vii vector<ii>
#define PQ priority_queue
#define minPQ priority_queue<int, vt<int>, greater<int>>
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define ER equal_range
#define ALL(x) begin(x), end(x)
#define SZ(x) ((int)(x).size())
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define FORE(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
#define UNI(c) c.resize(distance(c.begin(), unique(c.begin(), c.end())));
#define SORT_UNI(c)                                                            \
  (sort(c.begin(), c.end()),                                                   \
   c.resize(distance(c.begin(), unique(c.begin(), c.end()))))
#define POS(c, x) (lower_bound(c.begin(), c.end(), (x)) - c.begin())
#define max(a, b) ((a) < (b) ? b : a)
#define min(a, b) ((a) < (b) ? a : b)
#define setmin(a, b) a = min(a, b)
#define setmax(a, b) a = max(a, b)
#define MINMAX(x) minmax_element(begin(x), end(x))

// permanent constants
const long long INF64 = 1e18;
const int INF32 = 1e9;
const double EPS = static_cast<double>(1e-10);
const double PI = 2.0 * acos(0.0); // acos(-1.0L);
const int MOD = (int)1e9 + 7;      // 998244353
ll powMod(ll a, ll b) {
  ll res = 1;
  a %= MOD;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1)
      res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
ll invMod(ll a) {
  ll b = MOD, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  return (MOD + u) % MOD;
}
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

// random
mt19937 rnd(
    239); //(chrono::high_resolution_clock::now().time_since_epoch().count());

// Tricks:
//  Sort w Lambda Function: sort(begin(v), end(v), [] (int a, int b) { return a
//  > b; }); Sort vector cua bo ba so: struct MyStr {int a, b, c;};
//  vector<MyStr> v; ...fill v... sort(all(v), [](MyStr &A, MyStr &B) { return
//  tie(A.a, A.b, A.c) < tie(B.a, B.b, B.c); });
#define odd(x) ((x)&1)
#define even(x) !((x)&1)
// vec - vector, p(x) - predicate so that p(x) = true for all elements of some
// prefix of vector vec and false on all others. To find the first place where
// p(x) doesn't hold one can simply use:
#define FIRST_FAIL(vec, p) partition_point(ALL(vec), p) - vec.begin()
// partition_point(begin, end, p) returns the first ITERATOR it that p(*it) =
// false
#define isIn(vec, key) binary_search(ALL(vec), key)
#define toBool(num) !!(num)
#define bin(N, num) cerr << bitset<N>(num) << "\n"
// vector a: tinh XOR sum tu l den r:
#define XORsum(a, l, r)                                                        \
  accumulate(a.begin() + l - 1, a.begin() + r, 0,                              \
             [](int x, int y) { return x ^ y; })
// BASIC STRING: just like vector but allows to use a few string member
// functions/operators: operator+ and operator+=.:
/*  int n;    basic_string<int> a;
    cin >> n; FOR(i, 0, n) { int x;   cin >> x; a += x; }
    a += a; a = a.substr(n/2, n);
    cout << (a + a).find({1, 2, 1}) << '\n';
*/
// count(a.begin(), a.end(), val) - count how many val in an array a
vector<int> prefixSum(vector<int> &a) {
  vi b;
  std::partial_sum(a.begin(), a.end(), b.begin());
  return b;
}

int mostSignDigit(int n) {
  double k = log10(n);
  return (int)pow(10, k - floor(k));
}
int numOfDigits(int n) { return (int)floor(log10(n)) + 1; }
bool isPowerOfTwo(int x) {
  return x && (!(x & (x - 1)));
} // first x is for the case x=0

bool allPositive(vector<int> &a) {
  return all_of(begin(a), end(a), [](int x) { return x > 0; });
} // are all of the elements positive?
// CHECK there is all numbers in array is positive: bool ok=true; for(auto z: a)
// ok &= (z>0);
bool anyPositive(vector<int> &a) {
  return any_of(begin(a), end(a), [](int x) { return x > 0; });
} // is there at least one positive element?
bool nonePositive(vector<int> &a) {
  return none_of(begin(a), end(a), [](int x) { return x > 0; });
} // are none of the elements positive?

void FILEIO() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
// void FILEIO() {  freopen("berries.in", "r", stdin);  freopen("berries.out",
// "w", stdout); } // USACO

/****************** DEBUG *******************/
// Universal DEBUG: https://codeforces.com/blog/entry/68809
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
// void __print(long long x) {cerr << x;}
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "T" : "F"); }
template <typename T, typename V> void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T> void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x)
    cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V> void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v))
    cerr << ", ";
  _print(v...);
}

// #define ONLINE_JUDGE
#ifndef ONLINE_JUDGE
#define debug(x...)                                                            \
  cerr << "[" << #x << "] = [";                                                \
  _print(x)
#else
#define debug(x...)
#endif
//********************** Codes go here *********************/

//************ MODULAR ARITHMETIC LIB ***************************/
template <int MOD> struct Fp {
  long long val;
  constexpr Fp(long long v = 0) noexcept
      : val(v % MOD) { // dinh nghia Fp(v) = (v % MOD)
    if (val < 0)
      val += MOD;
  }
  constexpr int getmod() { return MOD; } // get MOD
  constexpr Fp operator-() const noexcept {
    return val ? MOD - val : 0;
  } // phep toan "-"
  constexpr Fp operator+(const Fp &r) const noexcept {
    return Fp(*this) += r;
  } // phep toan "+="
  constexpr Fp operator-(const Fp &r) const noexcept {
    return Fp(*this) -= r;
  } // phep toan "-="
  constexpr Fp operator*(const Fp &r) const noexcept {
    return Fp(*this) *= r;
  } // phep toan "*="
  constexpr Fp operator/(const Fp &r) const noexcept {
    return Fp(*this) /= r;
  } // phep toan "/="
  constexpr Fp &operator+=(const Fp &r) noexcept {
    val += r.val;
    if (val >= MOD)
      val -= MOD;
    return *this;
  }
  constexpr Fp &operator-=(const Fp &r) noexcept {
    val -= r.val;
    if (val < 0)
      val += MOD;
    return *this;
  }
  constexpr Fp &operator*=(const Fp &r) noexcept {
    val = val * r.val % MOD;
    return *this;
  }
  constexpr Fp &operator/=(
      const Fp &r) noexcept { // dinh nghia phep chia, dung thuat toan inverse
    long long a = r.val, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    val = val * u % MOD;
    if (val < 0)
      val += MOD;
    return *this;
  }
  constexpr bool operator==(const Fp &r) const noexcept {
    return this->val == r.val;
  }
  constexpr bool operator!=(const Fp &r) const noexcept {
    return this->val != r.val;
  }
  friend constexpr ostream &operator<<(ostream &os, const Fp<MOD> &x) noexcept {
    return os << x.val;
  }
  friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
    if (n == 0)
      return 1;
    auto t = modpow(a, n / 2);
    t = t * t;
    if (n & 1)
      t = t * a;
    return t;
  }
};

template <class T> struct BiCoef { // tinh C(k, n)
  vector<T> fact_, inv_, finv_;
  constexpr BiCoef() {}
  constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
    init(n);
  }
  constexpr void init(int n) noexcept {
    fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
    // int MOD = fact_[0].getmod(); // thuong trong code se co global const nay
    for (int i = 2; i < n; i++) {
      fact_[i] = fact_[i - 1] * i;          // i!
      inv_[i] = -inv_[MOD % i] * (MOD / i); // inverse cua so i
      finv_[i] = finv_[i - 1] * inv_[i];    // factorial inverse
    }
  }
  constexpr T C(int k, int n) const noexcept { // tinh C(k, n)
    if (n < k || n < 0 || k < 0)
      return 0;
    return fact_[n] * finv_[k] * finv_[n - k];
  }
  constexpr T fact(int n) const noexcept {
    if (n < 0)
      return 0;
    return fact_[n];
  }
  constexpr T inv(int n) const noexcept {
    if (n < 0)
      return 0;
    return inv_[n];
  }
  constexpr T finv(int n) const noexcept {
    if (n < 0)
      return 0;
    return finv_[n];
  }
};

// const int MOD = 1000000007;
using mint = Fp<MOD>; // mint = modular int
BiCoef<mint> bc;      // tao class de tinh C(k, n)
void preCalc() {
  bc.init(1000050);
} // precompute tat ca C(k,n) voi n = 500050, k = 1, n

void solve() {
  int k;
  string s;
  cin >> k >> s;
  int n = s.length();
  preCalc();
  mint sum = 0;
  FORE(x, 0, k) {
    mint a = powMod(26, x) * powMod(25, k - x) % MOD;
    sum += a * bc.C(k - x, k - x + n - 1);
    ;
  }
  cout << sum << endl;
}

signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  // FILEIO();
  //  auto beginProgram = chrono::steady_clock::now();

  // preCalc(); // any 1-time precalculation need to do before many test cases?

  // int t; cin >> t; FORE(i, 1, t)
  {
    // cout << "Case #" << i << ": "; // Google CodeJam style
    solve();
  }

  //  auto endProgram = chrono::steady_clock::now();
  //  cerr << chrono::duration_cast<chrono::milliseconds>(endProgram -
  //  beginProgram).count() << " ms" << endl;
  return 0;
}

Fixed submission:

// Problem : F - Strivore
// Contest : AtCoder - AtCoder Beginner Contest 171
// URL : https://atcoder.jp/contests/abc171/tasks/abc171_f
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

/*
We want to count the number of strings T of length |S|+K such that the given
string S is a subsequence of T. Let's fix T and try to build an algorithm to
check if S is a subsequence. Our algorithm if successful will return
a1,a2,...,a|S| such that Si=Tai. Let's build a in a way such that ai is minimum
possible for all i. Note that for each i<|S|, characters Tai+1,...,Tai+1−1 can't
be equal to Tai+1 (otherwise we can minimize ai+1), so for those characters we
have limited choices (only 25). Characters after a|S| can be anything (26
choices).

So, let's loop over how many characters will be after a|S|; let that number be
x. We have 26x choices for that suffix. For each of the other K−x characters, we
only have 25 choices, so we have 25K−x choices for them in total. Still, we have
to decide how to distribute these K−x characters. We have |S| slots for them (
before the first character of S, between the first and the second character,
...). Using stars and bars, we can distribute them in (|S|−1−k−x|S|−1) ways. So,
we will loop over x and add to the answer 25K−x∗26x∗(|S|−1−k−x|S|−1).
*/

/* ID: vqt
LANG: C++14
PROB: namenum
*/

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>
#include <unistd.h>
using namespace std;

// for PDBS use only:
//  #include <ext/pb_ds/assoc_container.hpp>
//  #include <ext/pb_ds/tree_policy.hpp>
//  using namespace __gnu_pbds;
//  template <typename T> using o_set = tree<T, null_type, greater<T>,
//  rb_tree_tag, tree_order_statistics_node_update>;
typedef long long ll;
// #define ordered_set tree<pair<ll,ll>, null_type,less<pair<ll,ll>>,
// rb_tree_tag,tree_order_statistics_node_update>

// defines
#define int long long
// #define double long double
#define vec vector
#define vi vector<int>
#define vvi vector<vector<int>>
#define ii pair<int, int>
#define F first
#define S second
#define MP make_pair
#define vii vector<ii>
#define PQ priority_queue
#define minPQ priority_queue<int, vt<int>, greater<int>>
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define ER equal_range
#define ALL(x) begin(x), end(x)
#define SZ(x) ((int)(x).size())
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define FORE(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
#define UNI(c) c.resize(distance(c.begin(), unique(c.begin(), c.end())));
#define SORT_UNI(c)                                                            \
  (sort(c.begin(), c.end()),                                                   \
   c.resize(distance(c.begin(), unique(c.begin(), c.end()))))
#define POS(c, x) (lower_bound(c.begin(), c.end(), (x)) - c.begin())
#define max(a, b) ((a) < (b) ? b : a)
#define min(a, b) ((a) < (b) ? a : b)
#define setmin(a, b) a = min(a, b)
#define setmax(a, b) a = max(a, b)
#define MINMAX(x) minmax_element(begin(x), end(x))

// permanent constants
const long long INF64 = 1e18;
const int INF32 = 1e9;
const double EPS = static_cast<double>(1e-10);
const double PI = 2.0 * acos(0.0); // acos(-1.0L);
const int MOD = (int)1e9 + 7;      // 998244353
ll powMod(ll a, ll b) {
  ll res = 1;
  a %= MOD;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1)
      res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
ll invMod(ll a) {
  ll b = MOD, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  return (MOD + u) % MOD;
}
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

// random
mt19937 rnd(
    239); //(chrono::high_resolution_clock::now().time_since_epoch().count());

// Tricks:
//  Sort w Lambda Function: sort(begin(v), end(v), [] (int a, int b) { return a
//  > b; }); Sort vector cua bo ba so: struct MyStr {int a, b, c;};
//  vector<MyStr> v; ...fill v... sort(all(v), [](MyStr &A, MyStr &B) { return
//  tie(A.a, A.b, A.c) < tie(B.a, B.b, B.c); });
#define odd(x) ((x)&1)
#define even(x) !((x)&1)
// vec - vector, p(x) - predicate so that p(x) = true for all elements of some
// prefix of vector vec and false on all others. To find the first place where
// p(x) doesn't hold one can simply use:
#define FIRST_FAIL(vec, p) partition_point(ALL(vec), p) - vec.begin()
// partition_point(begin, end, p) returns the first ITERATOR it that p(*it) =
// false
#define isIn(vec, key) binary_search(ALL(vec), key)
#define toBool(num) !!(num)
#define bin(N, num) cerr << bitset<N>(num) << "\n"
// vector a: tinh XOR sum tu l den r:
#define XORsum(a, l, r)                                                        \
  accumulate(a.begin() + l - 1, a.begin() + r, 0,                              \
             [](int x, int y) { return x ^ y; })
// BASIC STRING: just like vector but allows to use a few string member
// functions/operators: operator+ and operator+=.:
/*  int n;    basic_string<int> a;
    cin >> n; FOR(i, 0, n) { int x;   cin >> x; a += x; }
    a += a; a = a.substr(n/2, n);
    cout << (a + a).find({1, 2, 1}) << '\n';
*/
// count(a.begin(), a.end(), val) - count how many val in an array a
vector<int> prefixSum(vector<int> &a) {
  vi b;
  std::partial_sum(a.begin(), a.end(), b.begin());
  return b;
}

int mostSignDigit(int n) {
  double k = log10(n);
  return (int)pow(10, k - floor(k));
}
int numOfDigits(int n) { return (int)floor(log10(n)) + 1; }
bool isPowerOfTwo(int x) {
  return x && (!(x & (x - 1)));
} // first x is for the case x=0

bool allPositive(vector<int> &a) {
  return all_of(begin(a), end(a), [](int x) { return x > 0; });
} // are all of the elements positive?
// CHECK there is all numbers in array is positive: bool ok=true; for(auto z: a)
// ok &= (z>0);
bool anyPositive(vector<int> &a) {
  return any_of(begin(a), end(a), [](int x) { return x > 0; });
} // is there at least one positive element?
bool nonePositive(vector<int> &a) {
  return none_of(begin(a), end(a), [](int x) { return x > 0; });
} // are none of the elements positive?

void FILEIO() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
// void FILEIO() {  freopen("berries.in", "r", stdin);  freopen("berries.out",
// "w", stdout); } // USACO

/****************** DEBUG *******************/
// Universal DEBUG: https://codeforces.com/blog/entry/68809
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
// void __print(long long x) {cerr << x;}
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "T" : "F"); }
template <typename T, typename V> void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T> void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x)
    cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V> void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v))
    cerr << ", ";
  _print(v...);
}

// #define ONLINE_JUDGE
#ifndef ONLINE_JUDGE
#define debug(x...)                                                            \
  cerr << "[" << #x << "] = [";                                                \
  _print(x)
#else
#define debug(x...)
#endif
//********************** Codes go here *********************/

//************ MODULAR ARITHMETIC LIB ***************************/
template <int MOD> struct Fp {
  long long val;
  constexpr Fp(long long v = 0) noexcept
      : val(v % MOD) { // dinh nghia Fp(v) = (v % MOD)
    if (val < 0)
      val += MOD;
  }
  constexpr int getmod() { return MOD; } // get MOD
  constexpr Fp operator-() const noexcept {
    return val ? MOD - val : 0;
  } // phep toan "-"
  constexpr Fp operator+(const Fp &r) const noexcept {
    return Fp(*this) += r;
  } // phep toan "+="
  constexpr Fp operator-(const Fp &r) const noexcept {
    return Fp(*this) -= r;
  } // phep toan "-="
  constexpr Fp operator*(const Fp &r) const noexcept {
    return Fp(*this) *= r;
  } // phep toan "*="
  constexpr Fp operator/(const Fp &r) const noexcept {
    return Fp(*this) /= r;
  } // phep toan "/="
  constexpr Fp &operator+=(const Fp &r) noexcept {
    val += r.val;
    if (val >= MOD)
      val -= MOD;
    return *this;
  }
  constexpr Fp &operator-=(const Fp &r) noexcept {
    val -= r.val;
    if (val < 0)
      val += MOD;
    return *this;
  }
  constexpr Fp &operator*=(const Fp &r) noexcept {
    val = val * r.val % MOD;
    return *this;
  }
  constexpr Fp &operator/=(
      const Fp &r) noexcept { // dinh nghia phep chia, dung thuat toan inverse
    long long a = r.val, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    val = val * u % MOD;
    if (val < 0)
      val += MOD;
    return *this;
  }
  constexpr bool operator==(const Fp &r) const noexcept {
    return this->val == r.val;
  }
  constexpr bool operator!=(const Fp &r) const noexcept {
    return this->val != r.val;
  }
  friend constexpr ostream &operator<<(ostream &os, const Fp<MOD> &x) noexcept {
    return os << x.val;
  }
  friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
    if (n == 0)
      return 1;
    auto t = modpow(a, n / 2);
    t = t * t;
    if (n & 1)
      t = t * a;
    return t;
  }
};

template <class T> struct BiCoef { // tinh C(k, n)
  vector<T> fact_, inv_, finv_;
  constexpr BiCoef() {}
  constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
    init(n);
  }
  constexpr void init(int n) noexcept {
    fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
    // int MOD = fact_[0].getmod(); // thuong trong code se co global const nay
    for (int i = 2; i < n; i++) {
      fact_[i] = fact_[i - 1] * i;          // i!
      inv_[i] = -inv_[MOD % i] * (MOD / i); // inverse cua so i
      finv_[i] = finv_[i - 1] * inv_[i];    // factorial inverse
    }
  }
  constexpr T C(int k, int n) const noexcept { // tinh C(k, n)
    if (n < k || n < 0 || k < 0)
      return 0;
    return fact_[n] * finv_[k] * finv_[n - k];
  }
  constexpr T fact(int n) const noexcept {
    if (n < 0)
      return 0;
    return fact_[n];
  }
  constexpr T inv(int n) const noexcept {
    if (n < 0)
      return 0;
    return inv_[n];
  }
  constexpr T finv(int n) const noexcept {
    if (n < 0)
      return 0;
    return finv_[n];
  }
};

// const int MOD = 1000000007;
using mint = Fp<MOD>; // mint = modular int
BiCoef<mint> bc;      // tao class de tinh C(k, n)
void preCalc() {
  bc.init(2000050);
} // precompute tat ca C(k,n) voi n = 500050, k = 1, n

void solve() {
  int k;
  string s;
  cin >> k >> s;
  int n = s.length();
  preCalc();
  mint sum = 0;
  FORE(x, 0, k) {
    mint a = powMod(26, x) * powMod(25, k - x) % MOD;
    sum += a * bc.C(k - x, k - x + n - 1);
    ;
  }
  cout << sum << endl;
}

signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  // FILEIO();
  //  auto beginProgram = chrono::steady_clock::now();

  // preCalc(); // any 1-time precalculation need to do before many test cases?

  // int t; cin >> t; FORE(i, 1, t)
  {
    // cout << "Case #" << i << ": "; // Google CodeJam style
    solve();
  }

  //  auto endProgram = chrono::steady_clock::now();
  //  cerr << chrono::duration_cast<chrono::milliseconds>(endProgram -
  //  beginProgram).count() << " ms" << endl;
  return 0;
}