#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0;
  bool flg = false;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') flg = true;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
  return flg ? -x : x;
}
int mod;
inline int upd(const int &x) { return x + (x >> 31 & mod); }
inline void add(int &x, const int &y) { x = upd(x + y - mod); }
inline void iadd(int &x, const int &y) { x = upd(x - y); }
int qpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1LL * x * x % mod)
    if (y & 1) res = 1LL * res * x % mod;
  return res;
}
const int n = 400;
int fac[410], f[410], g[410], h[410][410];
void prework() {
  fac[0] = 1;
  for (int i(1); i <= int(n + 3); i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  for (int i(1); i <= int(n); i++) {
    g[i] = fac[i];
    for (int j(1); j <= int(i - 1); j++)
      iadd(g[i], 1LL * g[j] * fac[i - j] % mod);
  }
  h[0][0] = 1;
  for (int i(1); i <= int(n); i++) {
    for (int j(1); j <= int(i); j++) {
      for (int k(1); k <= int(i); k++) {
        add(h[i][j], 1LL * h[i - k][j - 1] * fac[k] % mod);
      }
    }
  }
  f[1] = 1, f[2] = 2, f[3] = 0;
  for (int i(4); i <= int(n); i++) {
    f[i] = fac[i];
    for (int j(1); j <= int(i - 1); j++)
      iadd(f[i], 2LL * g[j] * fac[i - j] % mod);
    for (int j(4); j <= int(i - 1); j++) iadd(f[i], 1LL * h[i][j] * f[j] % mod);
  }
}
void solve() { printf("%d\n", f[read()]); }
int main() {
  int T = read();
  mod = read();
  prework();
  for (; T--;) solve();
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int read() {
  int x = 0;
  bool flg = false;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') flg = true;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
  return flg ? -x : x;
}
int mod;
inline int upd(const int &x) { return x + (x >> 31 & mod); }
inline void add(int &x, const int &y) { x = upd(x + y - mod); }
inline void iadd(int &x, const int &y) { x = upd(x - y); }
int qpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1LL * x * x % mod)
    if (y & 1) res = 1LL * res * x % mod;
  return res;
}
const int n = 400;
int fac[410], f[410], g[410], h[410][410];
void prework() {
  fac[0] = 1;
  for (int i(1); i <= int(n + 3); i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  for (int i(1); i <= int(n); i++) {
    g[i] = fac[i];
    for (int j(1); j <= int(i - 1); j++)
      iadd(g[i], 1LL * g[j] * fac[i - j] % mod);
  }
  h[0][0] = 1;
  for (int i(1); i <= int(n); i++) {
    for (int j(1); j <= int(i); j++) {
      for (int k(1); k <= int(i); k++) {
        add(h[i][j], 1LL * h[i - k][j - 1] * fac[k] % mod);
      }
    }
  }
  f[1] = 1, f[2] = 2, f[3] = 0;
  for (int i(4); i <= int(n); i++) {
    f[i] = fac[i];
    for (int j(1); j <= int(i - 1); j++)
      iadd(f[i], 2LL * g[j] * fac[i - j] % mod);
    for (int j(4); j <= int(i - 1); j++) iadd(f[i], 1LL * h[i][j] * f[j] % mod);
  }
}
void solve() { printf("%d\n", f[read()]); }
int main() {
  int T = read();
  mod = read();
  prework();
  for (; T--;) solve();
  return 0;
}

```