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;
mt19937 rnd(228);
const int N = 404;
int mod;
int fact[N], d0[N], d[N], d2[N][N];
int main() {
  int qq;
  cin >> qq >> mod;
  fact[0] = 1;
  for (int i = 1; i < N; ++i) fact[i] = (1LL * fact[i - 1] * i) % mod;
  d[1] = 1;
  d0[1] = 1;
  for (int n = 2; n < N; ++n) {
    d0[n] = fact[n];
    for (int k = 1; k < n; ++k)
      d0[n] = (d0[n] - 1LL * d0[k] * fact[n - k]) % mod;
    if (d0[n] < 0) d0[n] += mod;
  }
  d2[0][0] = 1;
  for (int n = 1; n < N; ++n) {
    d2[n][1] = fact[n];
    for (int k = 2; k <= n; ++k)
      for (int x = 1; x < n; ++x)
        d2[n][k] = (d2[n][k] + 1LL * fact[x] * d2[n - x][k - 1]) % mod;
  }
  for (int n = 2; n < N; ++n) {
    int dz = 0;
    for (int x = 1; x < n - 1; ++x)
      for (int y = x; y < n - 1; ++y) {
        int c = fact[y - x + 1];
        c = (1LL * c * d0[x]) % mod;
        c = (1LL * c * d0[n - y - 1]) % mod;
        dz += c;
        if (dz >= mod) dz -= mod;
      }
    dz = (dz * 2) % mod;
    int cd = fact[n];
    cd = (cd - dz + mod) % mod;
    for (int k = 3; k < n; ++k) {
      cd = (cd - 1LL * d2[n][k] * d[k]) % mod;
    }
    if (n != 2) {
      for (int x = 1; x < n; ++x) cd = (cd - 1LL * d0[x] * d0[n - x]) % mod;
      for (int x = 1; x < n; ++x) cd = (cd - 1LL * d0[x] * d0[n - x]) % mod;
    }
    cd = (cd + mod) % mod;
    d[n] = cd;
  }
  while (qq--) {
    int x;
    cin >> x;
    cout << d[x] << "\n";
  }
}