#include <bits/stdc++.h>
using namespace std;
long long p;
long long ksm(long long a, long long b) {
  if (!b) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % p;
  if (b & 1) ns = ns * a % p;
  return ns;
}
long long jc[405], bjc[405];
long long ans[405][405];
long long dp[405];
long long ns[405];
int main() {
  int t;
  scanf("%d%lld", &t, &p);
  jc[0] = bjc[0] = 1;
  for (int i = 1; i < 405; i++)
    jc[i] = jc[i - 1] * i % p, bjc[i] = ksm(jc[i], p - 2);
  for (int i = 0; i < 405; i++)
    for (int j = 0; j < 405; j++) {
      if (i == 0)
        if (j == 0)
          ans[i][j] = 1;
        else
          ans[i][j] = 0;
      else {
        for (int k = 1; k <= i; k++)
          ans[i][j] = (ans[i][j] + jc[k] * ans[i - k][j - 1]) % p;
      }
    }
  ns[1] = 1;
  for (int i = 2; i < 405; i++) {
    ns[i] = jc[i];
    for (int j = 1; j < i; j++) ns[i] -= ns[j] * jc[i - j], ns[i] %= p;
  }
  dp[1] = 1;
  for (int i = 2; i < 405; i++) {
    dp[i] = jc[i];
    for (int x = 2; x < i; x++) {
      long long nans = -1;
      if (x >= 3) {
        nans *= ans[i][x], nans %= p;
        nans *= dp[x], nans %= p;
      } else
        nans *= 2 * (jc[i] - ns[i]);
      dp[i] += nans;
      dp[i] %= p;
    }
    if (dp[i] < 0) dp[i] += p;
  }
  for (int i = 0; i < t; i++) {
    int n;
    scanf("%d", &n);
    printf("%lld\n", dp[n]);
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
long long p;
long long ksm(long long a, long long b) {
  if (!b) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % p;
  if (b & 1) ns = ns * a % p;
  return ns;
}
long long jc[405], bjc[405];
long long ans[405][405];
long long dp[405];
long long ns[405];
int main() {
  int t;
  scanf("%d%lld", &t, &p);
  jc[0] = bjc[0] = 1;
  for (int i = 1; i < 405; i++)
    jc[i] = jc[i - 1] * i % p, bjc[i] = ksm(jc[i], p - 2);
  for (int i = 0; i < 405; i++)
    for (int j = 0; j < 405; j++) {
      if (i == 0)
        if (j == 0)
          ans[i][j] = 1;
        else
          ans[i][j] = 0;
      else {
        for (int k = 1; k <= i; k++)
          ans[i][j] = (ans[i][j] + jc[k] * ans[i - k][j - 1]) % p;
      }
    }
  ns[1] = 1;
  for (int i = 2; i < 405; i++) {
    ns[i] = jc[i];
    for (int j = 1; j < i; j++) ns[i] -= ns[j] * jc[i - j], ns[i] %= p;
  }
  dp[1] = 1;
  for (int i = 2; i < 405; i++) {
    dp[i] = jc[i];
    for (int x = 2; x < i; x++) {
      long long nans = -1;
      if (x >= 3) {
        nans *= ans[i][x], nans %= p;
        nans *= dp[x], nans %= p;
      } else
        nans *= 2 * (jc[i] - ns[i]);
      dp[i] += nans;
      dp[i] %= p;
    }
    if (dp[i] < 0) dp[i] += p;
  }
  for (int i = 0; i < t; i++) {
    int n;
    scanf("%d", &n);
    printf("%lld\n", dp[n]);
  }
  return 0;
}

```