#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int n, mod, fac[N], ifac[N];
inline void inc(int& x, int y) { x = x + y >= mod ? x + y - mod : x + y; }
inline int qpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    (y & 1) && (res = 1ll * res * x % mod);
  return res;
}
inline void pre() {
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[n] = qpow(fac[n], mod - 2);
  for (int i = n - 1; ~i; --i) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}
int f[N][N][30], ans;
inline void work(int a1) {
  memset(f, 0, sizeof f);
  int mx = n + 1 - a1;
  f[0][0][0] = 1;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= a1; ++j)
      for (int k = 0; k <= mx; ++k) {
        int& d = f[i][j][k];
        if (!d) continue;
        for (int ct = 1; ct <= n - i; ++ct) {
          if (i + ct > k + a1) break;
          if (i + ct == k + a1 && i) break;
          if (i + ct > a1 && k + a1 < i + ct + 1) break;
          int sum = j + ct * k;
          if (i + ct == n) {
            inc(ans, 1ll * d * ifac[ct] % mod);
            break;
          }
          for (int t = k + 1; t <= mx && sum + (n - ct - i) * t <= a1; ++t)
            inc(f[i + ct][sum][t], 1ll * d * ifac[ct] % mod);
        }
      }
}
int main() {
  scanf("%d%d", &n, &mod);
  pre();
  for (int i = max(1, n - 2 * (int)(sqrt(n + 0.01))); i <= n + 1; ++i) work(i);
  ans = 1ll * ans * fac[n] % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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], ifac[N];
inline void inc(int& x, int y) { x = x + y >= mod ? x + y - mod : x + y; }
inline int qpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    (y & 1) && (res = 1ll * res * x % mod);
  return res;
}
inline void pre() {
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[n] = qpow(fac[n], mod - 2);
  for (int i = n - 1; ~i; --i) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}
int f[N][N][30], ans;
inline void work(int a1) {
  memset(f, 0, sizeof f);
  int mx = n + 1 - a1;
  f[0][0][0] = 1;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= a1; ++j)
      for (int k = 0; k <= mx; ++k) {
        int& d = f[i][j][k];
        if (!d) continue;
        for (int ct = 1; ct <= n - i; ++ct) {
          if (i + ct > k + a1) break;
          if (i + ct == k + a1 && i) break;
          if (i + ct > a1 && k + a1 < i + ct + 1) break;
          int sum = j + ct * k;
          if (i + ct == n) {
            inc(ans, 1ll * d * ifac[ct] % mod);
            break;
          }
          for (int t = k + 1; t <= mx && sum + (n - ct - i) * t <= a1; ++t)
            inc(f[i + ct][sum][t], 1ll * d * ifac[ct] % mod);
        }
      }
}
int main() {
  scanf("%d%d", &n, &mod);
  pre();
  for (int i = max(1, n - 2 * (int)(sqrt(n + 0.01))); i <= n + 1; ++i) work(i);
  ans = 1ll * ans * fac[n] % mod;
  printf("%d\n", ans);
  return 0;
}

```