#include <bits/stdc++.h>
using namespace std;
const int kMaxN = 420;
int mod;
int fact[kMaxN];
int nopref[kMaxN];
int dp_fact[kMaxN][kMaxN];
int dp_nopref[kMaxN][kMaxN];
int ngood[kMaxN];
int nbad[kMaxN];
void MakeFactorials() {
  fact[0] = 1;
  for (int i = 1; i < kMaxN; ++i) {
    fact[i] = (long long)fact[i - 1] * i % mod;
  }
}
void MakeNoPref() {
  for (int n = 1; n < kMaxN; ++n) {
    nopref[n] = fact[n];
    for (int k = 1; k < n; ++k) {
      nopref[n] = (nopref[n] - (long long)nopref[k] * fact[n - k]) % mod;
      if (nopref[n] < 0) {
        nopref[n] += mod;
      }
    }
  }
}
void MakeProdSum(int src[kMaxN], int dest[kMaxN][kMaxN]) {
  dest[0][0] = 1;
  for (int cnt = 1; cnt < kMaxN; ++cnt) {
    for (int n = cnt; n < kMaxN; ++n) {
      for (int i = 1; i <= n; ++i) {
        dest[cnt][n] =
            (dest[cnt][n] + (long long)dest[cnt - 1][n - i] * src[i]) % mod;
      }
    }
  }
}
void CountGoodPerms() {
  ngood[1] = 1;
  ngood[2] = 2;
  nbad[3] = 6;
  for (int n = 4; n < kMaxN; ++n) {
    for (int k = 4; k < n; ++k) {
      nbad[n] = (nbad[n] + (long long)ngood[k] * dp_fact[k][n]) % mod;
    }
    for (int k = 2; k <= n; ++k) {
      nbad[n] = (nbad[n] + 2LL * dp_nopref[k][n]) % mod;
    }
    ngood[n] = (mod + fact[n] - nbad[n]) % mod;
  }
}
void Preproc() {
  MakeFactorials();
  MakeNoPref();
  MakeProdSum(fact, dp_fact);
  MakeProdSum(nopref, dp_nopref);
  CountGoodPerms();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t >> mod;
  Preproc();
  while (t--) {
    int n;
    cin >> n;
    cout << ngood[n] << "\n";
  }
}

### Prompt

Create a solution in Cpp for the following problem:

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 kMaxN = 420;
int mod;
int fact[kMaxN];
int nopref[kMaxN];
int dp_fact[kMaxN][kMaxN];
int dp_nopref[kMaxN][kMaxN];
int ngood[kMaxN];
int nbad[kMaxN];
void MakeFactorials() {
  fact[0] = 1;
  for (int i = 1; i < kMaxN; ++i) {
    fact[i] = (long long)fact[i - 1] * i % mod;
  }
}
void MakeNoPref() {
  for (int n = 1; n < kMaxN; ++n) {
    nopref[n] = fact[n];
    for (int k = 1; k < n; ++k) {
      nopref[n] = (nopref[n] - (long long)nopref[k] * fact[n - k]) % mod;
      if (nopref[n] < 0) {
        nopref[n] += mod;
      }
    }
  }
}
void MakeProdSum(int src[kMaxN], int dest[kMaxN][kMaxN]) {
  dest[0][0] = 1;
  for (int cnt = 1; cnt < kMaxN; ++cnt) {
    for (int n = cnt; n < kMaxN; ++n) {
      for (int i = 1; i <= n; ++i) {
        dest[cnt][n] =
            (dest[cnt][n] + (long long)dest[cnt - 1][n - i] * src[i]) % mod;
      }
    }
  }
}
void CountGoodPerms() {
  ngood[1] = 1;
  ngood[2] = 2;
  nbad[3] = 6;
  for (int n = 4; n < kMaxN; ++n) {
    for (int k = 4; k < n; ++k) {
      nbad[n] = (nbad[n] + (long long)ngood[k] * dp_fact[k][n]) % mod;
    }
    for (int k = 2; k <= n; ++k) {
      nbad[n] = (nbad[n] + 2LL * dp_nopref[k][n]) % mod;
    }
    ngood[n] = (mod + fact[n] - nbad[n]) % mod;
  }
}
void Preproc() {
  MakeFactorials();
  MakeNoPref();
  MakeProdSum(fact, dp_fact);
  MakeProdSum(nopref, dp_nopref);
  CountGoodPerms();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t >> mod;
  Preproc();
  while (t--) {
    int n;
    cin >> n;
    cout << ngood[n] << "\n";
  }
}

```