#include <bits/stdc++.h>
const int N = 205;
int MOD;
inline int mval(int x) { return x >= MOD ? x - MOD : x; }
inline void inc(int &x, int a) { x = mval(x + a); }
int n, fac[N], inv[N], f[N][N][N], ans;
inline int barret(unsigned long long c) {
  static __int128 x = -1ull / MOD;
  int r = c - (unsigned long long)((x * c) >> 64) * MOD;
  return mval(r);
}
int main() {
  scanf("%d%d", &n, &MOD);
  fac[0] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; ++i)
    inv[i] = MOD - 1ll * (MOD / i) * inv[MOD % i] % MOD;
  for (int i = 1; i <= n; ++i)
    fac[i] = 1ll * i * fac[i - 1] % MOD,
    inv[i] = 1ll * inv[i - 1] * inv[i] % MOD;
  f[n + 2][0][0] = 1;
  for (int i = n + 1; i; --i) {
    for (int j = 0; j < n; ++j)
      for (int s = 0; s <= n; ++s)
        if (f[i + 1][j][s]) {
          int val = f[i + 1][j][s];
          for (int t = 0; t + j <= n && s + t + j <= n; ++t)
            if (t + j >= n + 2 - i)
              inc(f[i][j + t][s + t + j], barret(1ll * val * inv[t]));
          if (s <= i) inc(ans, barret(1ll * inv[n - j] * val));
          if (j + i - 1 == n && i > 1 && s + j <= i - 1)
            inc(ans, barret(1ll * inv[n - j] * val));
        }
  }
  printf("%d\n", (1ll * ans * fac[n]) % MOD);
  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 N = 205;
int MOD;
inline int mval(int x) { return x >= MOD ? x - MOD : x; }
inline void inc(int &x, int a) { x = mval(x + a); }
int n, fac[N], inv[N], f[N][N][N], ans;
inline int barret(unsigned long long c) {
  static __int128 x = -1ull / MOD;
  int r = c - (unsigned long long)((x * c) >> 64) * MOD;
  return mval(r);
}
int main() {
  scanf("%d%d", &n, &MOD);
  fac[0] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; ++i)
    inv[i] = MOD - 1ll * (MOD / i) * inv[MOD % i] % MOD;
  for (int i = 1; i <= n; ++i)
    fac[i] = 1ll * i * fac[i - 1] % MOD,
    inv[i] = 1ll * inv[i - 1] * inv[i] % MOD;
  f[n + 2][0][0] = 1;
  for (int i = n + 1; i; --i) {
    for (int j = 0; j < n; ++j)
      for (int s = 0; s <= n; ++s)
        if (f[i + 1][j][s]) {
          int val = f[i + 1][j][s];
          for (int t = 0; t + j <= n && s + t + j <= n; ++t)
            if (t + j >= n + 2 - i)
              inc(f[i][j + t][s + t + j], barret(1ll * val * inv[t]));
          if (s <= i) inc(ans, barret(1ll * inv[n - j] * val));
          if (j + i - 1 == n && i > 1 && s + j <= i - 1)
            inc(ans, barret(1ll * inv[n - j] * val));
        }
  }
  printf("%d\n", (1ll * ans * fac[n]) % MOD);
  return 0;
}

```