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. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 410;
int n, q, p, t[N][N], fac[N], s[N][N];
inline int Qmi(int x, int y = p - 2) {
  int ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % p)
    if (y & 1) ans = 1ll * ans * x % p;
  return ans;
}
inline void Inverse() {
  for (int i = 1; i <= n; ++i) s[i][i] = 1;
  for (int i = 1; i <= n; ++i) {
    if (!t[i][i]) {
      for (int j = i + 1; j <= n; ++j)
        if (t[j][i]) {
          for (int k = 1; k <= n; ++k) {
            swap(t[i][k], t[j][k]);
            swap(s[i][k], s[j][k]);
          }
          break;
        }
      assert(t[i][i]);
    }
    int qu = Qmi(s[i][i]);
    for (int j = 1; j <= n; ++j) {
      t[i][j] = 1ll * t[i][j] * qu % p;
      s[i][j] = 1ll * s[i][j] * qu % p;
    }
    for (int j = i + 1; j <= n; ++j) {
      int qu = t[j][i];
      for (int k = 1; k <= n; ++k) {
        t[j][k] = (t[j][k] - 1ll * t[i][k] * qu) % p;
        s[j][k] = (s[j][k] - 1ll * s[i][k] * qu) % p;
      }
    }
  }
}
inline void Debug() {
  int ans[100][100];
  memset(ans, 0, sizeof ans);
  puts("Check");
  for (int i = 1; i <= n; ++i) t[i][1] = fac[i];
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= i; ++j)
      for (int k = 0; k <= i - j; ++k)
        t[i][j] = (t[i][j] + 1ll * fac[k + 1] * t[i - k - 1][j - 1]) % p;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      for (int k = 1; k <= n; ++k)
        ans[i][k] = (ans[i][k] + 1ll * t[i][j] * s[j][k]) % p;
  puts("ans[][]");
  for (int i = 1; i <= n; puts(""), ++i)
    for (int j = 1; j <= n; ++j) printf("%d ", ans[i][j]);
}
int main() {
  scanf("%d%d", &q, &p);
  n = 400;
  fac[0] = 1;
  for (int i = 1; i < N; ++i) fac[i] = 1ll * fac[i - 1] * i % p;
  for (int i = 1; i <= n; ++i) t[i][1] = fac[i];
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= i; ++j)
      for (int k = 0; k <= i - j; ++k)
        t[i][j] = (t[i][j] + 1ll * fac[k + 1] * t[i - k - 1][j - 1]) % p;
  Inverse();
  for (int i = 1, x; i <= q; ++i) {
    scanf("%d", &x);
    if (x == 2) {
      puts("2");
      continue;
    }
    int ans = (-s[x][1] + (~x & 1 ? -2 : 2)) % p;
    if (ans < 0) ans += p;
    printf("%d\n", ans);
  }
  return 0;
}