#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int n, mod;
int f[N][N][N], fac[N], ifac[N];
int add(int x, int y) { return x + y < mod ? x + y : x + y - mod; }
int power(int a, int n) {
  int tp = 1;
  while (n) {
    if (n & 1) tp = 1ll * tp * a % mod;
    a = 1ll * a * a % mod, n >>= 1;
  }
  return tp;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> mod;
  int ans = 2;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[0] = 1;
  for (int i = 1; i <= n; i++)
    ifac[i] = 1ll * ifac[i - 1] * power(i, mod - 2) % mod;
  for (int i = 1; i < n; i++) f[n + 1][i][0] = ifac[i];
  for (int i = n; i; i--) {
    for (int j = 1; j < n; j++)
      for (int k = 0; k + j <= i; k++) {
        int l = n - j;
        if (l <= i)
          ans = (ans + 1ll * f[i + 1][j][k] * ifac[l] % mod * fac[n]) % mod;
      }
    for (int j = n - i + 1; j < n; j++)
      for (int k = n - i; k + j <= i; k++)
        f[i][j][k + j] = add(f[i][j][k + j], f[i + 1][j][k]);
    for (int j = n - 1; j > n - i; j--)
      for (int l = 1; l + j < n; l++)
        for (int k = n + 1 - i; k <= i; k++)
          f[i][j + l][k] = (f[i][j + l][k] + 1ll * f[i][j][k] * ifac[l]) % mod;
  }
  cout << ans;
}

### 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;
int f[N][N][N], fac[N], ifac[N];
int add(int x, int y) { return x + y < mod ? x + y : x + y - mod; }
int power(int a, int n) {
  int tp = 1;
  while (n) {
    if (n & 1) tp = 1ll * tp * a % mod;
    a = 1ll * a * a % mod, n >>= 1;
  }
  return tp;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> mod;
  int ans = 2;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[0] = 1;
  for (int i = 1; i <= n; i++)
    ifac[i] = 1ll * ifac[i - 1] * power(i, mod - 2) % mod;
  for (int i = 1; i < n; i++) f[n + 1][i][0] = ifac[i];
  for (int i = n; i; i--) {
    for (int j = 1; j < n; j++)
      for (int k = 0; k + j <= i; k++) {
        int l = n - j;
        if (l <= i)
          ans = (ans + 1ll * f[i + 1][j][k] * ifac[l] % mod * fac[n]) % mod;
      }
    for (int j = n - i + 1; j < n; j++)
      for (int k = n - i; k + j <= i; k++)
        f[i][j][k + j] = add(f[i][j][k + j], f[i + 1][j][k]);
    for (int j = n - 1; j > n - i; j--)
      for (int l = 1; l + j < n; l++)
        for (int k = n + 1 - i; k <= i; k++)
          f[i][j + l][k] = (f[i][j + l][k] + 1ll * f[i][j][k] * ifac[l]) % mod;
  }
  cout << ans;
}

```