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;
void sub(int &a, int b, int mod) {
  a -= b;
  if (a < 0) a += mod;
}
void add(int &a, int b, int mod) {
  a += b;
  if (a >= mod) a -= mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t, p;
  cin >> t >> p;
  int n = 400;
  vector<int> fact(n + 1);
  vector<int> I(n + 1), A(n + 1);
  vector<vector<int>> B(n + 1, vector<int>(n + 1));
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = 1ll * fact[i - 1] * i % p;
  I[0] = 1;
  for (int i = 1; i <= n; i++) {
    I[i] = fact[i];
    for (int len = 1; len < i; len++) {
      sub(I[i], 1ll * I[len] * fact[i - len] % p, p);
    }
  }
  B[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      for (int last = 1; last <= i; last++) {
        add(B[i][j], 1ll * B[i - last][j - 1] * fact[last] % p, p);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    A[i] = fact[i];
    if (i > 2)
      for (int len = 1; len < i; len++)
        sub(A[i], 2ll * I[len] * fact[i - len] % p, p);
    for (int cnt = 3; cnt < i; cnt++)
      sub(A[i], 1ll * B[i][cnt] * A[cnt] % p, p);
  }
  while (t--) {
    int x;
    cin >> x;
    cout << A[x] << "\n";
  }
  return 0;
}