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;
long long get() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const long long N = 405;
long long P, A[N], I[N], B[N][N], fac[N], inv[N];
long long qpow(long long x, long long y) {
  long long res = 1;
  while (y) res = res * ((y & 1) ? x : 1) % P, x = x * x % P, y >>= 1;
  return res;
}
void init(long long n) {
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % P;
  inv[n] = qpow(fac[n], P - 2);
  for (long long i = n; i >= 1; i--) inv[i - 1] = inv[i] * i % P;
  I[1] = 1;
  for (long long i = 2; i <= n; i++) {
    I[i] = fac[i];
    for (long long j = 1; j < i; j++)
      I[i] = (I[i] - I[j] * fac[i - j] % P + P) % P;
  }
  B[0][0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= i; j++)
      for (long long k = 1; k <= i; k++)
        B[i][j] = (B[i][j] + B[i - k][j - 1] * fac[k] % P) % P;
  A[1] = 1, A[2] = 2, A[3] = 0;
  for (long long i = 4; i <= n; i++) {
    A[i] = fac[i];
    long long sum1 = 0, sum2 = 0;
    for (long long j = 1; j < i; j++) sum1 = (sum1 + I[j] * fac[i - j] % P) % P;
    A[i] = (A[i] - sum1 * 2 % P + P) % P;
    for (long long j = 4; j < i; j++) sum2 = (sum2 + B[i][j] * A[j] % P) % P;
    A[i] = (A[i] - sum2 + P) % P;
  }
}
signed main() {
  long long T = get();
  P = get();
  init(400);
  while (T--) printf("%d\n", A[get()]);
  return 0;
}