#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
void qread(int &x) {
  int neg = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = 10 * x + c - '0', c = getchar();
  x *= neg;
}
const int maxn = 405;
int test, n, mod;
int dp[maxn][maxn * 2], fac[maxn];
int quickpow(int x, int pow) {
  int res = 1;
  for (; pow; pow >>= 1) {
    if (pow & 1) res = 1LL * res * x % mod;
    x = 1LL * x * x % mod;
  }
  return res;
}
void work() {
  for (int(i) = 1; (i) <= 400; i++) dp[i][i + 400] = 1;
  for (int(i) = 1; (i) <= 400; i++) {
    int inv = quickpow(dp[i][i], mod - 2);
    for (int j = i + 1; j <= 2 * 400; j++)
      dp[i][j] = 1LL * dp[i][j] * inv % mod;
    dp[i][i] = 1;
    for (int j = i + 1; j <= 400; j++) {
      for (int k = i + 1; k <= 2 * 400; k++)
        dp[j][k] = (dp[j][k] - 1LL * dp[j][i] * dp[i][k] % mod + mod) % mod;
      dp[j][i] = 0;
    }
  }
}
int main() {
  scanf("%d%d", &test, &mod);
  fac[0] = 1;
  for (int(i) = 1; (i) <= 400; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  for (int(i) = 1; (i) <= 400; i++) {
    dp[i][1] = fac[i];
    for (int k = 2; k <= i; k++) {
      for (int m = 0; m <= i - k; m++)
        dp[i][k] =
            (dp[i][k] + 1LL * fac[m + 1] * dp[i - m - 1][k - 1] % mod) % mod;
    }
  }
  work();
  while (test--) {
    scanf("%d", &n);
    if (n == 2) {
      puts("2");
      continue;
    }
    int ans = ((mod - dp[n][401]) % mod + ((n & 1) ? 2 : -2) + mod) % mod;
    printf("%d\n", ans);
  }
  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>
#pragma GCC optimize(2)
using namespace std;
void qread(int &x) {
  int neg = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = 10 * x + c - '0', c = getchar();
  x *= neg;
}
const int maxn = 405;
int test, n, mod;
int dp[maxn][maxn * 2], fac[maxn];
int quickpow(int x, int pow) {
  int res = 1;
  for (; pow; pow >>= 1) {
    if (pow & 1) res = 1LL * res * x % mod;
    x = 1LL * x * x % mod;
  }
  return res;
}
void work() {
  for (int(i) = 1; (i) <= 400; i++) dp[i][i + 400] = 1;
  for (int(i) = 1; (i) <= 400; i++) {
    int inv = quickpow(dp[i][i], mod - 2);
    for (int j = i + 1; j <= 2 * 400; j++)
      dp[i][j] = 1LL * dp[i][j] * inv % mod;
    dp[i][i] = 1;
    for (int j = i + 1; j <= 400; j++) {
      for (int k = i + 1; k <= 2 * 400; k++)
        dp[j][k] = (dp[j][k] - 1LL * dp[j][i] * dp[i][k] % mod + mod) % mod;
      dp[j][i] = 0;
    }
  }
}
int main() {
  scanf("%d%d", &test, &mod);
  fac[0] = 1;
  for (int(i) = 1; (i) <= 400; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  for (int(i) = 1; (i) <= 400; i++) {
    dp[i][1] = fac[i];
    for (int k = 2; k <= i; k++) {
      for (int m = 0; m <= i - k; m++)
        dp[i][k] =
            (dp[i][k] + 1LL * fac[m + 1] * dp[i - m - 1][k - 1] % mod) % mod;
    }
  }
  work();
  while (test--) {
    scanf("%d", &n);
    if (n == 2) {
      puts("2");
      continue;
    }
    int ans = ((mod - dp[n][401]) % mod + ((n & 1) ? 2 : -2) + mod) % mod;
    printf("%d\n", ans);
  }
  return 0;
}

```