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 MAXN = 400;
int MOD;
int fac[405], inv[405];
int dp[405];
int pw[405][405];
void gen_dp() {
  fac[0] = 1;
  for (int i = 1; i <= (int)(MAXN); i++)
    fac[i] = 1LL * fac[i - 1] * (i + 1) % MOD;
  inv[1] = 1;
  for (int i = 2; i <= MAXN; i++)
    inv[i] = MOD - 1LL * MOD / i * inv[MOD % i] % MOD;
  dp[0] = 1;
  for (int i = 1; i <= (int)(MAXN); i++)
    for (int j = 1; j <= (int)(i); j++)
      dp[i] = (dp[i] - 1LL * dp[i - j] * fac[j] % MOD + MOD) % MOD;
  pw[0][0] = 1;
  for (int i = 1; i <= (int)(MAXN); i++)
    for (int j0 = 0; j0 < (int)(MAXN); j0++)
      for (int j1 = 0; j1 < (int)(MAXN); j1++)
        if (j0 + j1 < MAXN)
          pw[i][j0 + j1] =
              (pw[i][j0 + j1] + 1LL * pw[i - 1][j0] * dp[j1]) % MOD;
}
int solve(int n) {
  if (n == 2) return 2;
  return (MOD * 2 - 1LL * pw[n][n - 1] * inv[n] % MOD + (n & 1 ? 2 : -2)) % MOD;
}
int main() {
  int T, n;
  scanf("%d%d", &T, &MOD);
  gen_dp();
  while (T--) {
    scanf("%d", &n);
    printf("%d\n", solve(n));
  }
  return 0;
}