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;
template <class T>
using vv = vector<vector<T>>;
template <class T>
inline bool MX(T &l, const T &r) {
  return l < r ? l = r, 1 : 0;
}
template <class T>
inline bool MN(T &l, const T &r) {
  return l > r ? l = r, 1 : 0;
}
const long long MOD = 1e9 + 7;
int T;
int N;
long long P;
long long f[414];
long long g[414];
long long h[414];
long long l[414][414];
long long fac[414];
long long inv[414];
long long mypow(long long base, long long exp) {
  if (exp == 0) return 1 % P;
  long long res = mypow(base * base % P, exp / 2);
  if (exp % 2) res = res * base % P;
  return res;
}
long long nCr(long long n, long long r) {
  if (r < 0 || n < r) return 0;
  return fac[n] * inv[r] % P * inv[n - r] % P;
}
int main() {
  scanf("%d%lld", &T, &P);
  fac[0] = inv[0] = 1;
  for (int(i) = 1; (i) < (414); ++(i)) {
    fac[i] = fac[i - 1] * i % P;
    inv[i] = inv[i - 1] * mypow(i, P - 2) % P;
  }
  l[0][0] = 1;
  f[1] = 0;
  h[1] = 1;
  g[1] = 1;
  for (int(n) = 1; (n) < (414); ++(n)) {
    long long sub = 0;
    for (int(k) = 1; (k) < (n + 1); ++(k)) {
      if (k < n) sub = (sub + g[k] * fac[n - k] % P) % P;
      for (int(j) = 1; (j) < (n + 1); ++(j)) {
        l[n][k] += l[n - j][k - 1] * fac[j] % P;
        l[n][k] %= P;
      }
    }
    g[n] = (fac[n] + P - sub) % P;
    f[n] = 2 * sub % P;
    for (int(k) = 3; (k) < (n); ++(k)) {
      f[n] = (f[n] + l[n][k] * h[k] % P) % P;
    }
    h[n] = (fac[n] + P - f[n]) % P;
  }
  for (int(i) = 0; (i) < (T); ++(i)) {
    scanf("%d", &N);
    if (N == 2)
      printf("2\n");
    else
      printf("%lld\n", h[N]);
  }
}