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;
inline int read() {
  int w = 1, s = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return w * s;
}
int T, mod, I[100100], g[405][405], f[405], fac[405];
inline int mns(int a, int b) {
  a -= b;
  return a < 0 ? a + mod : a;
}
inline int pls(int a, int b) {
  a += b;
  return a >= mod ? a - mod : a;
}
inline void Solve() {
  fac[0] = 1;
  for (register int i = 1; i <= 400; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  I[1] = 1;
  for (register int i = 2; i <= 400; ++i) {
    I[i] = fac[i];
    for (register int j = 1; j <= i - 1; ++j) {
      I[i] = mns(I[i], 1ll * fac[i - j] * I[j] % mod);
    }
  }
  g[0][0] = 1;
  for (register int i = 0; i <= 400; ++i) {
    for (register int j = 0; j <= i; ++j) {
      for (register int k = 1; k <= 400; ++k) {
        if (i + k <= 400) {
          g[i + k][j + 1] = pls(g[i + k][j + 1], 1ll * g[i][j] * fac[k] % mod);
        }
      }
    }
  }
  f[1] = 1;
  for (register int i = 2; i <= 400; ++i) {
    f[i] = fac[i];
    for (register int j = 1; j <= i - 1; ++j) {
      f[i] = mns(f[i], 2ll * I[j] * fac[i - j] % mod);
    }
    for (register int j = 3; j <= i - 1; ++j) {
      f[i] = mns(f[i], 1ll * g[i][j] * f[j] % mod);
    }
  }
}
int main() {
  T = read(), mod = read();
  Solve();
  f[2] = 2;
  while (T--) {
    int x = read();
    cout << f[x] << "\n";
  }
  return 0;
}