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 *r) -> decltype(cerr << *r);
template <class c>
char dud(...);
struct muu {
  template <class c>
  muu &operator<<(const c &) {
    return *this;
  }
  muu &operator()() { return *this; }
};
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using pll = pair<ll, ll>;
using vll = vector<ll>;
const int nax = 407;
ll mod;
int t;
ll dob[nax];
ll gir[nax];
ll glo[nax];
ll sil[nax];
ll ple[nax][nax];
void dod(ll &a, ll b) { a = (a + b) % mod; }
void odj(ll &a, ll b) {
  a = (a - b) % mod;
  if (a < 0) a += mod;
}
int main() {
  scanf("%d%lld", &t, &mod);
  sil[0] = 1;
  for (int i = 1; i < nax; i++) sil[i] = (sil[i - 1] * i) % mod;
  dob[1] = 1;
  dob[2] = 2;
  gir[1] = 1;
  gir[2] = 1;
  ple[0][0] = 1;
  glo[1] = 1;
  glo[2] = 2;
  for (int j = 1; j < nax; j++)
    for (int l = 1; l < nax; l++)
      for (int k = 1; k <= l; k++) dod(ple[j][l], ple[j - 1][l - k] * sil[k]);
  for (int i = 3; i < nax; i++) {
    for (int j = 1; j < i; j++) {
      if (j > 2) {
        dod(glo[i], sil[i - j] * (sil[j] + mod - gir[j]));
      } else {
        dod(glo[i], sil[i - j]);
      }
    }
    gir[i] = glo[i];
    dod(dob[i], sil[i]);
    odj(dob[i], 2 * gir[i]);
    for (int j = 3; j < i; j++) odj(dob[i], dob[j] * ple[j][i]);
    dod(glo[i], gir[i] + dob[i]);
  }
  while (t--) {
    int n;
    scanf("%d", &n);
    printf("%lld\n", dob[n]);
  }
  return 0;
}