#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(long long a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(long long a) {
  write(a);
  puts("");
}
inline void wri(long long a) {
  write(a);
  putchar(' ');
}
const int N = 405;
long long dp[N][N], fac[N], ans[N], sum[N];
int main() {
  int T = read(), mod = read();
  int n = 400;
  for (int i = fac[0] = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    sum[i] = fac[i];
    for (int j = 1; j < i; j++) sum[i] = (sum[i] - sum[j] * fac[i - j]) % mod;
    for (int j = 1; j <= i; j++) {
      for (int k = 1; k <= i; k++) {
        dp[i][j] += dp[i - k][j - 1] * fac[k];
        if (!(k & 15)) dp[i][j] %= mod;
      }
      dp[i][j] %= mod;
    }
    ans[i] = fac[i];
    for (int j = 4; j < i; j++) ans[i] = (ans[i] - ans[j] * dp[i][j]) % mod;
    if (i > 2)
      for (int j = 1; j < i; j++)
        ans[i] = (ans[i] - 2 * sum[j] * fac[i - j]) % mod;
  }
  while (T--) {
    writeln((ans[read()] + mod) % mod);
  }
}

### Prompt

Generate a CPP solution to the following 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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(long long a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(long long a) {
  write(a);
  puts("");
}
inline void wri(long long a) {
  write(a);
  putchar(' ');
}
const int N = 405;
long long dp[N][N], fac[N], ans[N], sum[N];
int main() {
  int T = read(), mod = read();
  int n = 400;
  for (int i = fac[0] = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    sum[i] = fac[i];
    for (int j = 1; j < i; j++) sum[i] = (sum[i] - sum[j] * fac[i - j]) % mod;
    for (int j = 1; j <= i; j++) {
      for (int k = 1; k <= i; k++) {
        dp[i][j] += dp[i - k][j - 1] * fac[k];
        if (!(k & 15)) dp[i][j] %= mod;
      }
      dp[i][j] %= mod;
    }
    ans[i] = fac[i];
    for (int j = 4; j < i; j++) ans[i] = (ans[i] - ans[j] * dp[i][j]) % mod;
    if (i > 2)
      for (int j = 1; j < i; j++)
        ans[i] = (ans[i] - 2 * sum[j] * fac[i - j]) % mod;
  }
  while (T--) {
    writeln((ans[read()] + mod) % mod);
  }
}

```