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 main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, r;
  cin >> t >> r;
  int n = 402;
  long long f[n];
  f[0] = 1;
  for (long long i = 1; i < n; i++) f[i] = (f[i - 1] * i) % r;
  long long d[n], p[n], dp[n][n];
  p[1] = 1;
  d[1] = 1;
  for (long long i = 2; i < n; i++) {
    d[i] = f[i];
    for (long long j = 1; j < i; j++) d[i] = (d[i] - d[j] * f[i - j]) % r;
    if (d[i] < 0) d[i] += r;
  }
  dp[0][0] = 1;
  for (long long i = 1; i < n; i++) {
    dp[i][1] = f[i];
    for (long long j = 2; j <= i; j++)
      for (long long k = 1; k < i; k++)
        dp[i][j] = (dp[i][j] + f[k] * dp[i - k][j - 1]) % r;
  }
  for (long long i = 2; i < n; i++) {
    long long h = 0;
    for (long long j = 1; j < i - 1; j++) {
      for (long long k = j; k < i - 1; k++) {
        long long c = f[k - j + 1];
        c = (c * d[j]) % r;
        c = (c * d[i - k - 1]) % r;
        h += c;
        if (h >= r) h -= r;
      }
    }
    h = (h * 2) % r;
    long long s = f[i];
    s = (s - h + r) % r;
    for (long long j = 3; j < i; j++) {
      s = (s - dp[i][j] * p[j]) % r;
    }
    if (i != 2) {
      for (long long j = 1; j < i; j++) s = (s - d[j] * d[i - j]) % r;
      for (long long j = 1; j < i; j++) s = (s - d[j] * d[i - j]) % r;
    }
    s = (s + r) % r;
    p[i] = s;
  }
  while (t--) {
    long long a;
    cin >> a;
    cout << p[a] << endl;
  }
  return 0;
}