#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int n, mod, fac[N], inv[N], ans;
unordered_map<int, int> f[N];
int fpow_(int a, int b, int res = 1) {
  for (; b; b >>= 1, a = 1ll * a * a % mod) {
    if (b & 1) {
      res = 1ll * res * a % mod;
    }
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &mod);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
  }
  inv[n] = fpow_(fac[n], mod - 2);
  for (int i = n; i; i--) {
    inv[i - 1] = 1ll * inv[i] * i % mod;
  }
  for (int mn = n + 1; mn; mn--) {
    int res = 0;
    for (int i = 0; i <= n; i++) {
      f[i].clear();
    }
    f[0][n * (n + 1)] = 1;
    for (int i = n + 1; i > mn; i--) {
      for (int j = n - 1; j >= 0; j--) {
        int mx = mn * i;
        for (auto val : f[j]) {
          int lim = min(val.first, mx);
          for (int k = 1; j + k < n; k++) {
            lim -= i;
            if (lim < mn * (n - j - k)) {
              continue;
            }
            f[j + k][lim] = (f[j + k][lim] + 1ll * val.second * inv[k]) % mod;
          }
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (auto val : f[i]) {
        int lim = val.first;
        if (mn * (n - i) <= lim && (n - i) <= mn) {
          res = (res + 1ll * val.second * inv[n - i]) % mod;
        }
      }
    }
    ans = (ans + 1ll * res * fac[n] % mod) % mod;
  }
  printf("%d\n", (ans % mod + mod) % mod);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
const int N = 205;
int n, mod, fac[N], inv[N], ans;
unordered_map<int, int> f[N];
int fpow_(int a, int b, int res = 1) {
  for (; b; b >>= 1, a = 1ll * a * a % mod) {
    if (b & 1) {
      res = 1ll * res * a % mod;
    }
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &mod);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
  }
  inv[n] = fpow_(fac[n], mod - 2);
  for (int i = n; i; i--) {
    inv[i - 1] = 1ll * inv[i] * i % mod;
  }
  for (int mn = n + 1; mn; mn--) {
    int res = 0;
    for (int i = 0; i <= n; i++) {
      f[i].clear();
    }
    f[0][n * (n + 1)] = 1;
    for (int i = n + 1; i > mn; i--) {
      for (int j = n - 1; j >= 0; j--) {
        int mx = mn * i;
        for (auto val : f[j]) {
          int lim = min(val.first, mx);
          for (int k = 1; j + k < n; k++) {
            lim -= i;
            if (lim < mn * (n - j - k)) {
              continue;
            }
            f[j + k][lim] = (f[j + k][lim] + 1ll * val.second * inv[k]) % mod;
          }
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (auto val : f[i]) {
        int lim = val.first;
        if (mn * (n - i) <= lim && (n - i) <= mn) {
          res = (res + 1ll * val.second * inv[n - i]) % mod;
        }
      }
    }
    ans = (ans + 1ll * res * fac[n] % mod) % mod;
  }
  printf("%d\n", (ans % mod + mod) % mod);
}

```