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 N = 405;
int n, mod, I[N], a[N], b[N], f[N], cs[N][N], fac[N], ifac[N], inv[N];
void init(int x) {
  fac[0] = ifac[0] = inv[1] = 1;
  for (int i = (2); i <= (x); ++i)
    inv[i] = (long long)inv[mod % i] * (mod - mod / i) % mod;
  for (int i = (1); i <= (x); ++i)
    fac[i] = (long long)fac[i - 1] * i % mod,
    ifac[i] = (long long)ifac[i - 1] * inv[i] % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T, n = 400;
  cin >> T >> mod, init(n);
  cs[0][0] = 1;
  for (int i = (1); i <= (n); ++i)
    for (int j = (0); j <= (n); ++j)
      for (int k = (1); k <= (j); ++k)
        (cs[i][j] += (long long)fac[k] * cs[i - 1][j - k] % mod) %= mod;
  for (int i = (1); i <= (n); ++i) {
    for (int j = (1); j <= (i - 1); ++j)
      (I[i] += (long long)I[j] * fac[i - j] % mod) %= mod;
    I[i] = (fac[i] + mod - I[i]) % mod;
  }
  for (int i = (1); i <= (n); ++i)
    a[i] = (fac[i] - I[i] + mod) % mod, a[i] = (long long)a[i] * 2 % mod;
  for (int i = (1); i <= (n); ++i) b[i] = (fac[i] - a[i] + mod) % mod;
  f[1] = 1;
  for (int i = (2); i <= (n); ++i) {
    f[i] = b[i];
    for (int j = (2); j <= (i - 1); ++j)
      (f[i] += mod - (long long)f[j] * cs[j][i] % mod) %= mod;
  }
  while (T--) {
    int x;
    cin >> x;
    if (x == 2)
      cout << 2 << "\n";
    else
      cout << f[x] << "\n";
  }
  return 0;
}