#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void ckmax(T& x, T y) {
  x = (y > x ? y : x);
}
template <typename T>
inline void ckmin(T& x, T y) {
  x = (y < x ? y : x);
}
const int MAXN = 400;
int MOD;
inline int mod1(int x) { return x < MOD ? x : x - MOD; }
inline int mod2(int x) { return x < 0 ? x + MOD : x; }
inline void add(int& x, int y) { x = mod1(x + y); }
inline void sub(int& x, int y) { x = mod2(x - y); }
int fac[MAXN + 5];
int h[MAXN + 5];
int g[MAXN + 5][MAXN + 5];
int f[MAXN + 5];
void init(int n) {
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = (long long)fac[i - 1] * i % MOD;
  }
  h[1] = 1;
  for (int i = 2; i <= n; ++i) {
    h[i] = fac[i];
    for (int j = 1; j < i; ++j) {
      sub(h[i], (long long)h[j] * fac[i - j] % MOD);
    }
  }
  g[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= i; ++j) {
      for (int k = 1; k <= i; ++k) {
        add(g[i][j], (long long)g[i - k][j - 1] * fac[k] % MOD);
      }
    }
  }
  f[1] = 1;
  f[2] = 2;
  for (int i = 3; i <= n; ++i) {
    f[i] = 0;
    for (int j = 1; j < i; ++j) {
      add(f[i], (long long)h[j] * fac[i - j] % MOD);
    }
    f[i] = mod1(f[i] * 2);
    for (int j = 3; j < i; ++j) {
      add(f[i], (long long)g[i][j] * f[j] % MOD);
    }
    f[i] = mod2(fac[i] - f[i]);
  }
}
int main() {
  int T;
  cin >> T;
  cin >> MOD;
  init(MAXN);
  while (T--) {
    int n;
    cin >> n;
    cout << f[n] << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
template <typename T>
inline void ckmax(T& x, T y) {
  x = (y > x ? y : x);
}
template <typename T>
inline void ckmin(T& x, T y) {
  x = (y < x ? y : x);
}
const int MAXN = 400;
int MOD;
inline int mod1(int x) { return x < MOD ? x : x - MOD; }
inline int mod2(int x) { return x < 0 ? x + MOD : x; }
inline void add(int& x, int y) { x = mod1(x + y); }
inline void sub(int& x, int y) { x = mod2(x - y); }
int fac[MAXN + 5];
int h[MAXN + 5];
int g[MAXN + 5][MAXN + 5];
int f[MAXN + 5];
void init(int n) {
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = (long long)fac[i - 1] * i % MOD;
  }
  h[1] = 1;
  for (int i = 2; i <= n; ++i) {
    h[i] = fac[i];
    for (int j = 1; j < i; ++j) {
      sub(h[i], (long long)h[j] * fac[i - j] % MOD);
    }
  }
  g[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= i; ++j) {
      for (int k = 1; k <= i; ++k) {
        add(g[i][j], (long long)g[i - k][j - 1] * fac[k] % MOD);
      }
    }
  }
  f[1] = 1;
  f[2] = 2;
  for (int i = 3; i <= n; ++i) {
    f[i] = 0;
    for (int j = 1; j < i; ++j) {
      add(f[i], (long long)h[j] * fac[i - j] % MOD);
    }
    f[i] = mod1(f[i] * 2);
    for (int j = 3; j < i; ++j) {
      add(f[i], (long long)g[i][j] * f[j] % MOD);
    }
    f[i] = mod2(fac[i] - f[i]);
  }
}
int main() {
  int T;
  cin >> T;
  cin >> MOD;
  init(MAXN);
  while (T--) {
    int n;
    cin >> n;
    cout << f[n] << endl;
  }
  return 0;
}

```