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>
#pragma GCC optimize("O3")
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
const int kMaxN = 420;
int t, p;
int nops[kMaxN];
int bad[kMaxN];
int good[kMaxN];
int fact[kMaxN];
int dp_nops[kMaxN][kMaxN];
int dp_fact[kMaxN][kMaxN];
void Preproc() {
  fact[0] = 1;
  for (int i = 1; i < kMaxN; ++i) {
    fact[i] = (ll)fact[i - 1] * i % p;
  }
  for (int n = 1; n < kMaxN; ++n) {
    nops[n] = fact[n];
    for (int k = 1; k < n; ++k) {
      nops[n] = (nops[n] - (ll)nops[k] * fact[n - k]) % p;
      if (nops[n] < 0) {
        nops[n] += p;
      }
    }
  }
  dp_nops[0][0] = 1;
  for (int cnt = 1; cnt < kMaxN; ++cnt) {
    for (int n = cnt; n < kMaxN; ++n) {
      for (int i = 1; i <= n; ++i) {
        dp_nops[cnt][n] =
            (dp_nops[cnt][n] + (ll)dp_nops[cnt - 1][n - i] * nops[i]) % p;
      }
    }
  }
  dp_fact[0][0] = 1;
  for (int cnt = 1; cnt < kMaxN; ++cnt) {
    for (int n = cnt; n < kMaxN; ++n) {
      for (int i = 1; i <= n; ++i) {
        dp_fact[cnt][n] =
            (dp_fact[cnt][n] + (ll)dp_fact[cnt - 1][n - i] * fact[i]) % p;
      }
    }
  }
  good[1] = 1;
  bad[2] = 2;
  for (int n = 3; n < kMaxN; ++n) {
    int order_factor = 0;
    for (int cnt = 2; cnt <= n; ++cnt) {
      order_factor = (order_factor + dp_nops[cnt][n]) % p;
    }
    order_factor = (order_factor * 2) % p;
    int unorder_factor = 0;
    for (int k = 2; k < n; ++k) {
      unorder_factor = (unorder_factor + (ll)good[k] * dp_fact[k][n]) % p;
    }
    bad[n] = (order_factor + unorder_factor) % p;
    good[n] = (p + fact[n] - bad[n]) % p;
  }
  good[2] = 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(11);
  cerr << fixed << setprecision(6);
  cin >> t >> p;
  Preproc();
  while (t--) {
    int n;
    cin >> n;
    cout << good[n] << "\n";
  }
}