#include <bits/stdc++.h>
const int Maxn = 200;
int n, Mod;
int f[Maxn + 5][Maxn + 5][Maxn + 5];
int inv_f[Maxn + 5], fac[Maxn + 5], inv[Maxn + 5];
int main() {
  scanf("%d%d", &n, &Mod);
  inv[1] = 1;
  for (int i = 2; i <= n; i++) {
    inv[i] = 1ll * (Mod - Mod / i) * inv[Mod % i] % Mod;
  }
  fac[0] = inv_f[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % Mod;
    inv_f[i] = 1ll * inv_f[i - 1] * inv[i] % Mod;
  }
  for (int i = 1; i < n; i++) {
    f[n + 1][i][0] = inv_f[i];
  }
  int ans = 2;
  for (int i = n; i > 0; i--) {
    for (int j = 1; j < n; j++) {
      for (int k = 0; k + j <= i; k++) {
        if (n - j <= i) {
          ans =
              (ans + 1ll * f[i + 1][j][k] * inv_f[n - j] % Mod * fac[n]) % Mod;
        }
      }
    }
    for (int j = n - i + 1; j < n; j++) {
      for (int k = n - i; j + k <= i; k++) {
        f[i][j][j + k] = (f[i][j][j + k] + f[i + 1][j][k]) % Mod;
      }
    }
    for (int j = n - 1; j > n - i; j--) {
      for (int l = 1; j + l < n; l++) {
        for (int k = n + 1 - i; k <= i; k++) {
          f[i][j + l][k] = (f[i][j + l][k] + 1ll * f[i][j][k] * inv_f[l]) % Mod;
        }
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

A sequence of integers b_1, b_2, …, b_m is called good if max(b_1, b_2, …, b_m) ⋅ min(b_1, b_2, …, b_m) ≥ b_1 + b_2 + … + b_m.

A sequence of integers a_1, a_2, …, a_n is called perfect if every non-empty subsequence of a is good.

YouKn0wWho has two integers n and M, M is prime. Help him find the number, modulo M, of perfect sequences a_1, a_2, …, a_n such that 1 ≤ a_i ≤ n + 1 for each integer i from 1 to n.

A sequence d is a subsequence of a sequence c if d can be obtained from c by deletion of several (possibly, zero or all) elements.

Input

The first and only line of the input contains two space-separated integers n and M (1 ≤ n ≤ 200; 10^8 ≤ M ≤ 10^9). It is guaranteed that M is prime.

Output

Print a single integer — the number of perfect sequences modulo M.

Examples

Input


2 998244353


Output


4


Input


4 100000007


Output


32


Input


69 999999937


Output


456886663

Note

In the first test case, the perfect sequences are [2, 2], [2, 3], [3, 2] and [3, 3].

In the second test case, some of the perfect sequences are [3, 4, 3, 5], [4, 5, 4, 4], [4, 5, 5, 5] etc. One example of a sequence which is not perfect is [2, 3, 3, 4], because, for example, the subsequence [2, 3, 4] is not an good as 2 ⋅ 4 < 2 + 3 + 4.

### Response

```cpp
#include <bits/stdc++.h>
const int Maxn = 200;
int n, Mod;
int f[Maxn + 5][Maxn + 5][Maxn + 5];
int inv_f[Maxn + 5], fac[Maxn + 5], inv[Maxn + 5];
int main() {
  scanf("%d%d", &n, &Mod);
  inv[1] = 1;
  for (int i = 2; i <= n; i++) {
    inv[i] = 1ll * (Mod - Mod / i) * inv[Mod % i] % Mod;
  }
  fac[0] = inv_f[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % Mod;
    inv_f[i] = 1ll * inv_f[i - 1] * inv[i] % Mod;
  }
  for (int i = 1; i < n; i++) {
    f[n + 1][i][0] = inv_f[i];
  }
  int ans = 2;
  for (int i = n; i > 0; i--) {
    for (int j = 1; j < n; j++) {
      for (int k = 0; k + j <= i; k++) {
        if (n - j <= i) {
          ans =
              (ans + 1ll * f[i + 1][j][k] * inv_f[n - j] % Mod * fac[n]) % Mod;
        }
      }
    }
    for (int j = n - i + 1; j < n; j++) {
      for (int k = n - i; j + k <= i; k++) {
        f[i][j][j + k] = (f[i][j][j + k] + f[i + 1][j][k]) % Mod;
      }
    }
    for (int j = n - 1; j > n - i; j--) {
      for (int l = 1; j + l < n; l++) {
        for (int k = n + 1 - i; k <= i; k++) {
          f[i][j + l][k] = (f[i][j + l][k] + 1ll * f[i][j][k] * inv_f[l]) % Mod;
        }
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```