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 = 405;
int f[N], s[N][N], dp[N], cp[N], x[N], mo;
void add(int &x, int y) { (x += y) >= mo && (x -= mo); }
int main() {
  ios::sync_with_stdio(0);
  int ncase;
  cin >> ncase >> mo;
  dp[1] = s[1][1] = cp[1] = x[1] = f[1] = 1;
  for (int i = 2; i < N; i++) {
    f[i] = f[i - 1] * 1LL * i % mo;
    dp[i] = f[i];
    for (int j = 2; j < i; j++) {
      for (int k = 1; k < i; k++)
        add(s[i][j], 1LL * f[k] * s[i - k][j - 1] % mo);
      if (j == 2) {
        for (int k = 1; k < i; k++) add(x[i], 1LL * f[k] * cp[i - k] % mo);
      } else
        add(dp[i], mo - 1LL * s[i][j] * dp[j] % mo);
    }
    add(dp[i], mo - 2LL * x[i] % mo);
    s[i][1] = f[i];
    s[i][i] = 1;
    cp[i] = f[i];
    for (int j = 1; j < i; j++) add(cp[i], mo - 1LL * f[j] * cp[i - j] % mo);
  }
  while (ncase--) {
    int n;
    cin >> n;
    printf("%d\n", dp[n]);
  }
  return 0;
}