#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long rand(long long x, long long y) {
  return (rng() % (y + 1 - x)) + x;
}
string inline to_string(char c) {
  string second(1, c);
  return second;
}
template <typename T>
inline T gcd(T a, T b) {
  return a == 0 ? b : gcd(b % a, a);
}
long long int t, mod;
vector<long long int> first(401, 0);
long long int I_[406], ways_[406][406];
long long int I(long long int n) {
  if (~I_[n]) return I_[n];
  long long int ans = first[n];
  for (long long int i = (1); i <= (long long int)(n - 1); ++i) {
    ans -= I(i) * first[n - i] % mod, ans += mod, ans %= mod;
  }
  return I_[n] = ans;
}
long long int ways(long long int n, long long int k) {
  if (n < k || n < 0 || k < 0) return 0;
  if (k == 0 && n == 0) return 1;
  if (~ways_[n][k]) return ways_[n][k];
  long long int ans = 0;
  for (long long int i = (1); i <= (long long int)(n); ++i) {
    ans += ways(n - i, k - 1) * first[i] % mod, ans %= mod;
  }
  return ways_[n][k] = ans;
}
int main() {
  memset((I_), -1, sizeof((I_))), memset((ways_), -1, sizeof((ways_)));
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> t >> mod;
  first[0] = first[1] = 1;
  for (long long int i = (2); i <= (long long int)(400); ++i)
    first[i] = first[i - 1] * i % mod;
  vector<long long int> ans(406, 0);
  ans[1] = 1;
  for (long long int i = (2); i <= (long long int)(400); ++i) {
    ans[i] = first[i];
    for (long long int j = (3); j <= (long long int)(i - 1); ++j) {
      ans[i] -= ways(i, j) * ans[j] % mod, ans[i] += mod, ans[i] %= mod;
    }
    if (i ^ 2)
      for (long long int j = (1); j <= (long long int)(i - 1); ++j)
        ans[i] -= 2 * I(j) % mod * first[i - j] % mod, ans[i] += mod,
            ans[i] %= mod;
  }
  while (t--) {
    long long int n;
    cin >> n;
    cout << ans[n] << '\n';
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long rand(long long x, long long y) {
  return (rng() % (y + 1 - x)) + x;
}
string inline to_string(char c) {
  string second(1, c);
  return second;
}
template <typename T>
inline T gcd(T a, T b) {
  return a == 0 ? b : gcd(b % a, a);
}
long long int t, mod;
vector<long long int> first(401, 0);
long long int I_[406], ways_[406][406];
long long int I(long long int n) {
  if (~I_[n]) return I_[n];
  long long int ans = first[n];
  for (long long int i = (1); i <= (long long int)(n - 1); ++i) {
    ans -= I(i) * first[n - i] % mod, ans += mod, ans %= mod;
  }
  return I_[n] = ans;
}
long long int ways(long long int n, long long int k) {
  if (n < k || n < 0 || k < 0) return 0;
  if (k == 0 && n == 0) return 1;
  if (~ways_[n][k]) return ways_[n][k];
  long long int ans = 0;
  for (long long int i = (1); i <= (long long int)(n); ++i) {
    ans += ways(n - i, k - 1) * first[i] % mod, ans %= mod;
  }
  return ways_[n][k] = ans;
}
int main() {
  memset((I_), -1, sizeof((I_))), memset((ways_), -1, sizeof((ways_)));
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> t >> mod;
  first[0] = first[1] = 1;
  for (long long int i = (2); i <= (long long int)(400); ++i)
    first[i] = first[i - 1] * i % mod;
  vector<long long int> ans(406, 0);
  ans[1] = 1;
  for (long long int i = (2); i <= (long long int)(400); ++i) {
    ans[i] = first[i];
    for (long long int j = (3); j <= (long long int)(i - 1); ++j) {
      ans[i] -= ways(i, j) * ans[j] % mod, ans[i] += mod, ans[i] %= mod;
    }
    if (i ^ 2)
      for (long long int j = (1); j <= (long long int)(i - 1); ++j)
        ans[i] -= 2 * I(j) % mod * first[i - j] % mod, ans[i] += mod,
            ans[i] %= mod;
  }
  while (t--) {
    long long int n;
    cin >> n;
    cout << ans[n] << '\n';
  }
}

```