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;
long long dp[405][405], S[405], f[405], mod;
int main() {
  int T, n;
  cin >> T >> mod;
  int N = 400;
  f[0] = 1;
  for (long long i = 1; i <= N; ++i) f[i] = i * f[i - 1] % mod;
  for (int i = 1; i <= N; ++i) dp[1][i] = f[i];
  for (int i = 2; i <= N; ++i)
    for (int j = i; j <= N; ++j) {
      long long sum = 0;
      for (long long k = 1; k <= j; ++k)
        sum = (sum + f[k] * dp[i - 1][j - k] % mod) % mod;
      dp[i][j] = sum;
    }
  S[1] = 1;
  for (int i = 2; i <= N; ++i) {
    long long sum = f[i];
    for (int k = 2; k < i; ++k) {
      sum = (sum - S[k] * dp[k][i] % mod) % mod;
    }
    S[i] = sum;
  }
  for (int i = 3; i <= N; ++i) {
    if (i % 2 == 0)
      S[i] = (S[i] - 2) % mod;
    else
      S[i] = (S[i] + 2) % mod;
  }
  while (T--) {
    scanf("%d", &n);
    printf("%lld\n", (S[n] + mod) % mod);
  }
}