#include <bits/stdc++.h>
using namespace std;
int n, mod, ans;
int dp[205][205][205], fac[205], ifac[205];
int qpow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return ans;
}
void init(int n) {
  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 >= 0; --i) {
    ifac[i] = 1LL * ifac[i + 1] * (i + 1) % mod;
  }
}
int a1;
int dfs(int i, int j, int k) {
  if (!k) return !j ? ifac[i] : 0;
  if (n - i < n + 1 - a1 - k) return 0;
  if (~dp[i][j][k]) return dp[i][j][k];
  int &tmp = dp[i][j][k];
  tmp = 0;
  for (int cnt = 0; k * cnt <= j && cnt <= i; ++cnt) {
    tmp = (tmp + 1LL * ifac[cnt] * dfs(i - cnt, j - k * cnt, k - 1)) % mod;
  }
  return tmp;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> mod;
  init(n);
  for (a1 = max(1, n - 25); a1 <= n + 1; ++a1) {
    memset(dp, -1, sizeof(dp));
    for (int sum = 1; sum <= a1; ++sum) {
      ans = (ans + dfs(n, sum, n + 1 - a1)) % mod;
    }
  }
  cout << (1LL * ans * fac[n] + 1) % mod << '\n';
  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;
int n, mod, ans;
int dp[205][205][205], fac[205], ifac[205];
int qpow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return ans;
}
void init(int n) {
  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 >= 0; --i) {
    ifac[i] = 1LL * ifac[i + 1] * (i + 1) % mod;
  }
}
int a1;
int dfs(int i, int j, int k) {
  if (!k) return !j ? ifac[i] : 0;
  if (n - i < n + 1 - a1 - k) return 0;
  if (~dp[i][j][k]) return dp[i][j][k];
  int &tmp = dp[i][j][k];
  tmp = 0;
  for (int cnt = 0; k * cnt <= j && cnt <= i; ++cnt) {
    tmp = (tmp + 1LL * ifac[cnt] * dfs(i - cnt, j - k * cnt, k - 1)) % mod;
  }
  return tmp;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> mod;
  init(n);
  for (a1 = max(1, n - 25); a1 <= n + 1; ++a1) {
    memset(dp, -1, sizeof(dp));
    for (int sum = 1; sum <= a1; ++sum) {
      ans = (ans + dfs(n, sum, n + 1 - a1)) % mod;
    }
  }
  cout << (1LL * ans * fac[n] + 1) % mod << '\n';
  return 0;
}

```