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;
int n, mod;
const int maxn = 410;
int ans[maxn];
int dp[maxn];
int fac[maxn];
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
}
int ho[maxn];
int DP[maxn][maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q >> mod;
  dp[1] = fac[1] = 1;
  for (int i = 2; i <= 400; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  for (int i = 2; i <= 400; i++) {
    dp[i] = fac[i];
    for (int j = 1; j < i; j++) add(dp[i], -1LL * dp[j] * fac[i - j] % mod);
  }
  for (int i = 4; i <= 400; i++) {
    for (int j = 1; j < i; j++) add(ho[i], 1LL * dp[j] * fac[i - j] % mod);
    add(ho[i], ho[i]);
  }
  ans[4] = 2;
  DP[4][4] = 1;
  DP[0][0] = 1;
  for (int i = 1; i <= 400; i++) {
    DP[i][i] = 1;
    for (int j = 1; j < i; j++) {
      for (int k = 1; k <= i - j + 1; k++)
        add(DP[i][j], 1LL * fac[k] * DP[i - k][j - 1] % mod);
    }
  }
  for (int i = 5; i <= 400; i++) {
    ans[i] = fac[i];
    add(ans[i], -ho[i]);
    for (int j = 4; j < i; j++) add(ans[i], -1LL * DP[i][j] * ans[j] % mod);
  }
  while (q--) {
    int x;
    cin >> x;
    if (x == 1)
      cout << 1 << '\n';
    else if (x == 2)
      cout << 2 << '\n';
    else if (x == 3)
      cout << 0 << '\n';
    else
      cout << ans[x] << '\n';
  }
}