#include <bits/stdc++.h>
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
int vv[202 + 1], ff[202 + 1], gg[202 + 1];
void init(int md) {
  int i;
  ff[0] = gg[0] = 1;
  for (i = 1; i <= 202; i++) {
    vv[i] = i == 1 ? 1 : (long long)vv[i - md % i] * (md / i + 1) % md;
    ff[i] = (long long)ff[i - 1] * i % md;
    gg[i] = (long long)gg[i - 1] * vv[i] % md;
  }
}
int main() {
  static int dp[202 + 1][202 + 1];
  int n, md, a, b, k, l, s, ans;
  scanf("%d%d", &n, &md), init(md);
  ans = 0;
  for (a = 1; a <= n + 1; a++) {
    for (k = 0; k <= n; k++) memset(dp[k], 0, (a + 1) * sizeof *dp[k]);
    dp[0][0] = 1;
    for (b = a; b <= n + 1; b++) {
      for (k = n; k >= 0; k--)
        for (s = 0; s + (n - k) * (b - a) <= a && s + k * a <= a * (b - 1);
             s++) {
          int x = dp[k][s], k_, s_, lower, upper;
          if (x == 0) continue;
          lower = max(1, s + (n - k) * (b + 1 - a) - a);
          upper = min(n - k, (a * b - s - k * a) / b);
          for (l = lower; l <= upper; l++) {
            k_ = k + l, s_ = s + l * (b - a);
            dp[k_][s_] = (dp[k_][s_] + (long long)dp[k][s] * gg[l]) % md;
          }
        }
      if (b == a) dp[0][0] = 0;
    }
    for (s = 0; s <= a; s++) ans = (ans + (long long)dp[n][s] * ff[n]) % md;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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>
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
int vv[202 + 1], ff[202 + 1], gg[202 + 1];
void init(int md) {
  int i;
  ff[0] = gg[0] = 1;
  for (i = 1; i <= 202; i++) {
    vv[i] = i == 1 ? 1 : (long long)vv[i - md % i] * (md / i + 1) % md;
    ff[i] = (long long)ff[i - 1] * i % md;
    gg[i] = (long long)gg[i - 1] * vv[i] % md;
  }
}
int main() {
  static int dp[202 + 1][202 + 1];
  int n, md, a, b, k, l, s, ans;
  scanf("%d%d", &n, &md), init(md);
  ans = 0;
  for (a = 1; a <= n + 1; a++) {
    for (k = 0; k <= n; k++) memset(dp[k], 0, (a + 1) * sizeof *dp[k]);
    dp[0][0] = 1;
    for (b = a; b <= n + 1; b++) {
      for (k = n; k >= 0; k--)
        for (s = 0; s + (n - k) * (b - a) <= a && s + k * a <= a * (b - 1);
             s++) {
          int x = dp[k][s], k_, s_, lower, upper;
          if (x == 0) continue;
          lower = max(1, s + (n - k) * (b + 1 - a) - a);
          upper = min(n - k, (a * b - s - k * a) / b);
          for (l = lower; l <= upper; l++) {
            k_ = k + l, s_ = s + l * (b - a);
            dp[k_][s_] = (dp[k_][s_] + (long long)dp[k][s] * gg[l]) % md;
          }
        }
      if (b == a) dp[0][0] = 0;
    }
    for (s = 0; s <= a; s++) ans = (ans + (long long)dp[n][s] * ff[n]) % md;
  }
  printf("%d\n", ans);
  return 0;
}

```