#include <bits/stdc++.h>
int n, MOD, fact[205], ifact[205], dp[205][205][205], ans;
int power(int A, int B) {
  int res = 1;
  while (B) {
    if (B & 1) res = 1ll * res * A % MOD;
    A = 1ll * A * A % MOD;
    B >>= 1;
  }
  return res;
}
void ps(int &A, int B) { A = (A + B) % MOD; }
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin >> n >> MOD;
  fact[0] = 1;
  for (long long i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i % MOD;
  ifact[n] = power(fact[n], MOD - 2);
  for (long long i = n; i; --i) ifact[i - 1] = ifact[i] * i % MOD;
  for (int a1 = n + 1; a1 >= n - 2 * std::sqrt(n); --a1) {
    for (int i = 1; i <= std::min(n, a1); ++i)
      dp[i][0][std::max(i + 2, a1 + 1)] = ifact[i];
    for (int i = 1; i <= n; ++i)
      for (int j = 0; j <= a1; ++j)
        for (int k = a1 + 1; k <= n + 1; ++k)
          for (int l = 0;
               dp[i][j][k] && k >= i + l + 1 && j + l * (k - a1) <= a1; ++l)
            ps(dp[i + l][j + l * (k - a1)][std::max(i + l + 2, k + 1)],
               1ll * dp[i][j][k] * ifact[l] % MOD);
    for (int i = 0; i <= a1; ++i)
      ps(ans, 1ll * dp[n][i][n + 2] * fact[n] % MOD);
    memset(dp, 0, sizeof(dp));
  }
  std::cout << ans << std::endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 n, MOD, fact[205], ifact[205], dp[205][205][205], ans;
int power(int A, int B) {
  int res = 1;
  while (B) {
    if (B & 1) res = 1ll * res * A % MOD;
    A = 1ll * A * A % MOD;
    B >>= 1;
  }
  return res;
}
void ps(int &A, int B) { A = (A + B) % MOD; }
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin >> n >> MOD;
  fact[0] = 1;
  for (long long i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i % MOD;
  ifact[n] = power(fact[n], MOD - 2);
  for (long long i = n; i; --i) ifact[i - 1] = ifact[i] * i % MOD;
  for (int a1 = n + 1; a1 >= n - 2 * std::sqrt(n); --a1) {
    for (int i = 1; i <= std::min(n, a1); ++i)
      dp[i][0][std::max(i + 2, a1 + 1)] = ifact[i];
    for (int i = 1; i <= n; ++i)
      for (int j = 0; j <= a1; ++j)
        for (int k = a1 + 1; k <= n + 1; ++k)
          for (int l = 0;
               dp[i][j][k] && k >= i + l + 1 && j + l * (k - a1) <= a1; ++l)
            ps(dp[i + l][j + l * (k - a1)][std::max(i + l + 2, k + 1)],
               1ll * dp[i][j][k] * ifact[l] % MOD);
    for (int i = 0; i <= a1; ++i)
      ps(ans, 1ll * dp[n][i][n + 2] * fact[n] % MOD);
    memset(dp, 0, sizeof(dp));
  }
  std::cout << ans << std::endl;
  return 0;
}

```