#include <bits/stdc++.h>
int mod;
using namespace std;
int n, K;
int fac[210], ifac[210];
int f[210][210][210];
int ksm(int a, int k) {
  int res = 1;
  for (; k; k >>= 1, a = 1ll * a * a % mod)
    if (k & 1) res = 1ll * res * a % mod;
  return res;
}
void qmo(int &x) { x += (x >> 31) & mod; }
int work(int a1) {
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; i++) f[0][i][0] = ifac[i];
  for (int i = 1; i <= n + 1; i++)
    for (int j = n - n / i; j <= n && j - i + 1 <= a1; j++) {
      for (int k = 0; k <= n; k++)
        for (int l = 0; l <= j && l <= k / i; l++) {
          f[i][j][k] =
              (f[i][j][k] + 1ll * f[i - 1][j - l][k - i * l] * ifac[l]) % mod;
        }
    }
  int res = (a1 >= n ? ifac[n] : 0);
  for (int i = 1; i <= a1; i++) qmo(res += f[n + 1 - a1][n][i] - mod);
  return res;
}
int main() {
  scanf("%d%d", &n, &mod);
  int res = 0;
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= n; i++)
    fac[i] = 1ll * i * fac[i - 1] % mod, ifac[i] = ksm(fac[i], mod - 2),
    assert(1ll * fac[i] * ifac[i] % mod == 1);
  for (int a1 = n - 17; a1 <= n + 1; a1++) qmo(res += work(a1) - mod);
  printf("%d\n", 1ll * res * fac[n] % mod);
}

### 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 mod;
using namespace std;
int n, K;
int fac[210], ifac[210];
int f[210][210][210];
int ksm(int a, int k) {
  int res = 1;
  for (; k; k >>= 1, a = 1ll * a * a % mod)
    if (k & 1) res = 1ll * res * a % mod;
  return res;
}
void qmo(int &x) { x += (x >> 31) & mod; }
int work(int a1) {
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; i++) f[0][i][0] = ifac[i];
  for (int i = 1; i <= n + 1; i++)
    for (int j = n - n / i; j <= n && j - i + 1 <= a1; j++) {
      for (int k = 0; k <= n; k++)
        for (int l = 0; l <= j && l <= k / i; l++) {
          f[i][j][k] =
              (f[i][j][k] + 1ll * f[i - 1][j - l][k - i * l] * ifac[l]) % mod;
        }
    }
  int res = (a1 >= n ? ifac[n] : 0);
  for (int i = 1; i <= a1; i++) qmo(res += f[n + 1 - a1][n][i] - mod);
  return res;
}
int main() {
  scanf("%d%d", &n, &mod);
  int res = 0;
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= n; i++)
    fac[i] = 1ll * i * fac[i - 1] % mod, ifac[i] = ksm(fac[i], mod - 2),
    assert(1ll * fac[i] * ifac[i] % mod == 1);
  for (int a1 = n - 17; a1 <= n + 1; a1++) qmo(res += work(a1) - mod);
  printf("%d\n", 1ll * res * fac[n] % mod);
}

```