#include <bits/stdc++.h>
using namespace std;
int const N = 400;
int md, fc[N + 1], dp[N + 1][N + 1], an[N + 1], sad[N + 1];
void ad(int &x, int y) {
  if ((x += y) >= md) x -= md;
}
void sb(int &x, int y) {
  if ((x -= y) < 0) x += md;
}
void solve() {
  fc[0] = 1;
  for (int i = 1; i < (int)(N + 1); ++i) fc[i] = (long long)fc[i - 1] * i % md;
  dp[0][0] = 1;
  for (int i = 0; i < (int)(N); ++i)
    for (int j = i; j < (int)(N); ++j)
      for (int k = 1; k < (int)(N - j + 1); ++k)
        ad(dp[i + 1][j + k], (long long)dp[i][j] * fc[k] % md);
  for (int i = 1; i < (int)(N + 1); ++i) {
    sad[i] = fc[i];
    for (int j = 1; j < (int)(i); ++j)
      sb(sad[i], (long long)sad[j] * fc[i - j] % md);
  }
  for (int i = 1; i < (int)(N + 1); ++i) an[i] = fc[i];
  for (int i = 3; i < (int)(N + 1); ++i) {
    for (int j = 1; j < (int)(i); ++j) {
      int s = fc[j];
      for (int k = 1; k < (int)(j); ++k)
        sb(s, (long long)sad[k] * fc[j - k] % md);
      sb(an[i], 2ll * s * fc[i - j] % md);
    }
    for (int j = 3; j < (int)(i); ++j)
      sb(an[i], (long long)an[j] * dp[j][i] % md);
  }
}
int main() {
  int t;
  scanf("%d%d", &t, &md);
  solve();
  while (t--) {
    int t;
    scanf("%d", &t);
    printf("%d\n", an[t]);
  }
}

### Prompt

Develop a solution in cpp 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;
int const N = 400;
int md, fc[N + 1], dp[N + 1][N + 1], an[N + 1], sad[N + 1];
void ad(int &x, int y) {
  if ((x += y) >= md) x -= md;
}
void sb(int &x, int y) {
  if ((x -= y) < 0) x += md;
}
void solve() {
  fc[0] = 1;
  for (int i = 1; i < (int)(N + 1); ++i) fc[i] = (long long)fc[i - 1] * i % md;
  dp[0][0] = 1;
  for (int i = 0; i < (int)(N); ++i)
    for (int j = i; j < (int)(N); ++j)
      for (int k = 1; k < (int)(N - j + 1); ++k)
        ad(dp[i + 1][j + k], (long long)dp[i][j] * fc[k] % md);
  for (int i = 1; i < (int)(N + 1); ++i) {
    sad[i] = fc[i];
    for (int j = 1; j < (int)(i); ++j)
      sb(sad[i], (long long)sad[j] * fc[i - j] % md);
  }
  for (int i = 1; i < (int)(N + 1); ++i) an[i] = fc[i];
  for (int i = 3; i < (int)(N + 1); ++i) {
    for (int j = 1; j < (int)(i); ++j) {
      int s = fc[j];
      for (int k = 1; k < (int)(j); ++k)
        sb(s, (long long)sad[k] * fc[j - k] % md);
      sb(an[i], 2ll * s * fc[i - j] % md);
    }
    for (int j = 3; j < (int)(i); ++j)
      sb(an[i], (long long)an[j] * dp[j][i] % md);
  }
}
int main() {
  int t;
  scanf("%d%d", &t, &md);
  solve();
  while (t--) {
    int t;
    scanf("%d", &t);
    printf("%d\n", an[t]);
  }
}

```