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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = (long long)INF * INF;
const int MAX = 407;
int mod;
int add(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
int sub(int a, int b) { return a - b >= 0 ? a - b : a - b + mod; }
int mult(int a, int b) { return (long long)a * b % mod; }
int fact[MAX];
int dp1[MAX], dp2[MAX][MAX];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t >> mod;
  fact[0] = 1;
  for (int i = (1); i < (MAX); ++i) fact[i] = mult(fact[i - 1], i);
  dp2[0][0] = 1;
  for (int i = (1); i < (MAX); ++i)
    for (int j = (1); j < (MAX); ++j)
      for (int k = (1); k < (i + 1); ++k)
        dp2[i][j] = add(dp2[i][j], mult(dp2[i - k][j - 1], fact[k]));
  for (int i = (1); i < (MAX); ++i) {
    dp1[i] = fact[i];
    for (int j = (2); j < (i); ++j)
      dp1[i] = sub(dp1[i], mult(dp2[i][j], dp1[j]));
  }
  while (t--) {
    int n;
    cin >> n;
    int ans = dp1[n];
    if (n > 2) {
      if (n & 1)
        ans = add(ans, 2);
      else
        ans = sub(ans, 2);
    }
    cout << ans << "\n";
  }
  return 0;
}