#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long inf64 = 1e18;
const int N = 405;
int mod;
inline int _mul(int a, int b) { return (1ll * a * b) % mod; }
inline int _sum(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
inline void add_sum(int &x, int y) {
  if ((x += y) >= mod) {
    x -= mod;
  }
}
inline void add_sub(int &x, int y) {
  if ((x -= y) < 0) {
    x += mod;
  }
}
inline int _sub(int a, int b) { return a >= b ? a - b : a - b + mod; }
inline void add_mul(int &x, int y) { x = _mul(x, y); }
inline int _binpow(int x, int pw) {
  int res = 1;
  int tmp = x;
  while (pw > 0) {
    if (pw & 1) {
      add_mul(res, tmp);
    }
    pw >>= 1;
    add_mul(tmp, tmp);
  }
  return res;
}
int fact[2 * N];
int rfact[2 * N];
int F[N];
int A[N];
int B[N][N];
int CNK(int n, int k) {
  if (n < k) return 0;
  return _mul(fact[n], _mul(rfact[n - k], rfact[k]));
}
void precalc() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = _mul(fact[i - 1], i);
  }
  rfact[N - 1] = _binpow(fact[N - 1], mod - 2);
  for (int i = N - 2; i >= 0; i--) {
    rfact[i] = _mul(rfact[i + 1], i + 1);
  }
  F[0] = 1;
  A[0] = 1;
  B[0][0] = 1;
  for (int n = 1; n < N; n++) {
    A[n] = fact[n];
    for (int k = 1; k < n; k++) {
      add_sub(A[n], _mul(fact[k], A[n - k]));
    }
    for (int k = 1; k <= n; k++) {
      for (int t = 1; t <= n; t++) {
        add_sum(B[n][k], _mul(fact[t], B[n - t][k - 1]));
      }
    }
    F[n] = fact[n];
    int c1 = 0, c2 = 0;
    for (int k = 1; k < n; k++) {
      add_sum(c1, _mul(fact[k], A[n - k]));
    }
    for (int k = 3; k < n; k++) {
      add_sum(c2, _mul(B[n][k], F[k]));
    }
    add_sub(F[n], _mul(2, c1));
    add_sub(F[n], c2);
  }
  F[2] = 2;
}
int main() {
  int tt;
  cin >> tt >> mod;
  precalc();
  for (int ti = 0; ti < tt; ti++) {
    int n;
    cin >> n;
    cout << F[n] << "\n";
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

Consider a permutation p_1, p_2, ... p_n of integers from 1 to n. We call a sub-segment p_l, p_{l+1}, ..., p_{r-1}, p_{r} of the permutation an interval if it is a reordering of some set of consecutive integers. For example, the permutation (6,7,1,8,5,3,2,4) has the intervals (6,7), (5,3,2,4), (3,2), and others. 

Each permutation has some trivial intervals — the full permutation itself and every single element. We call a permutation interval-free if it does not have non-trivial intervals. In other words, interval-free permutation does not have intervals of length between 2 and n - 1 inclusive. 

Your task is to count the number of interval-free permutations of length n modulo prime number p. 

Input

In the first line of the input there are two integers t (1 ≤ t ≤ 400) and p (10^8 ≤ p ≤ 10^9) — the number of test cases to solve and the prime modulo. In each of the next t lines there is one integer n (1 ≤ n ≤ 400) — the length of the permutation. 

Output

For each of t test cases print a single integer — the number of interval-free permutations modulo p. 

Examples

Input


4 998244353
1
4
5
9


Output


1
2
6
28146


Input


1 437122297
20


Output


67777575

Note

For n = 1 the only permutation is interval-free. For n = 4 two interval-free permutations are (2,4,1,3) and (3,1,4,2). For n = 5 — (2,4,1,5,3), (2,5,3,1,4), (3,1,5,2,4), (3,5,1,4,2), (4,1,3,5,2), and (4,2,5,1,3). We will not list all 28146 for n = 9, but for example (4,7,9,5,1,8,2,6,3), (2,4,6,1,9,7,3,8,5), (3,6,9,4,1,5,8,2,7), and (8,4,9,1,3,6,2,7,5) are interval-free.

The exact value for n = 20 is 264111424634864638. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long inf64 = 1e18;
const int N = 405;
int mod;
inline int _mul(int a, int b) { return (1ll * a * b) % mod; }
inline int _sum(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
inline void add_sum(int &x, int y) {
  if ((x += y) >= mod) {
    x -= mod;
  }
}
inline void add_sub(int &x, int y) {
  if ((x -= y) < 0) {
    x += mod;
  }
}
inline int _sub(int a, int b) { return a >= b ? a - b : a - b + mod; }
inline void add_mul(int &x, int y) { x = _mul(x, y); }
inline int _binpow(int x, int pw) {
  int res = 1;
  int tmp = x;
  while (pw > 0) {
    if (pw & 1) {
      add_mul(res, tmp);
    }
    pw >>= 1;
    add_mul(tmp, tmp);
  }
  return res;
}
int fact[2 * N];
int rfact[2 * N];
int F[N];
int A[N];
int B[N][N];
int CNK(int n, int k) {
  if (n < k) return 0;
  return _mul(fact[n], _mul(rfact[n - k], rfact[k]));
}
void precalc() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = _mul(fact[i - 1], i);
  }
  rfact[N - 1] = _binpow(fact[N - 1], mod - 2);
  for (int i = N - 2; i >= 0; i--) {
    rfact[i] = _mul(rfact[i + 1], i + 1);
  }
  F[0] = 1;
  A[0] = 1;
  B[0][0] = 1;
  for (int n = 1; n < N; n++) {
    A[n] = fact[n];
    for (int k = 1; k < n; k++) {
      add_sub(A[n], _mul(fact[k], A[n - k]));
    }
    for (int k = 1; k <= n; k++) {
      for (int t = 1; t <= n; t++) {
        add_sum(B[n][k], _mul(fact[t], B[n - t][k - 1]));
      }
    }
    F[n] = fact[n];
    int c1 = 0, c2 = 0;
    for (int k = 1; k < n; k++) {
      add_sum(c1, _mul(fact[k], A[n - k]));
    }
    for (int k = 3; k < n; k++) {
      add_sum(c2, _mul(B[n][k], F[k]));
    }
    add_sub(F[n], _mul(2, c1));
    add_sub(F[n], c2);
  }
  F[2] = 2;
}
int main() {
  int tt;
  cin >> tt >> mod;
  precalc();
  for (int ti = 0; ti < tt; ti++) {
    int n;
    cin >> n;
    cout << F[n] << "\n";
  }
  return 0;
}

```