#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int mod;
int power(long long n, long long k) {
  int ans = 1 % mod;
  n %= mod;
  if (n < 0) n += mod;
  while (k) {
    if (k & 1) ans = (long long)ans * n % mod;
    n = (long long)n * n % mod;
    k >>= 1;
  }
  return ans;
}
int dp[N][N][40], fac[N], ifac[N];
int vis[N][N][40];
int n, a1;
int yo(int i, int sum, int k) {
  if (i == n) return fac[n];
  if (k == 0) return 1LL * fac[n] * ifac[n - i] % mod;
  if (vis[i][sum][k] == a1) return dp[i][sum][k];
  vis[i][sum][k] = a1;
  int &ans = dp[i][sum][k];
  ans = 0;
  int r = (a1 - sum) / k;
  for (int cnt = r; cnt >= 0; cnt--) {
    if (k > 1 and i + cnt < n - a1 + 1 - k + 1) continue;
    ans += 1LL * yo(i + cnt, sum + cnt * k, k - 1) * ifac[cnt] % mod;
    ans %= mod;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> mod;
  fac[0] = 1;
  ifac[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = 1LL * fac[i - 1] * i % mod;
    ifac[i] = power(fac[i], mod - 2);
  }
  int ans = 0;
  int lim = 2 * sqrt(n) + 1;
  for (a1 = max(1, n - lim); a1 <= n; a1++) {
    ans += yo(0, 0, n + 1 - a1);
    ans %= mod;
  }
  ans %= mod;
  cout << ans << '\n';
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
const int N = 205;
int mod;
int power(long long n, long long k) {
  int ans = 1 % mod;
  n %= mod;
  if (n < 0) n += mod;
  while (k) {
    if (k & 1) ans = (long long)ans * n % mod;
    n = (long long)n * n % mod;
    k >>= 1;
  }
  return ans;
}
int dp[N][N][40], fac[N], ifac[N];
int vis[N][N][40];
int n, a1;
int yo(int i, int sum, int k) {
  if (i == n) return fac[n];
  if (k == 0) return 1LL * fac[n] * ifac[n - i] % mod;
  if (vis[i][sum][k] == a1) return dp[i][sum][k];
  vis[i][sum][k] = a1;
  int &ans = dp[i][sum][k];
  ans = 0;
  int r = (a1 - sum) / k;
  for (int cnt = r; cnt >= 0; cnt--) {
    if (k > 1 and i + cnt < n - a1 + 1 - k + 1) continue;
    ans += 1LL * yo(i + cnt, sum + cnt * k, k - 1) * ifac[cnt] % mod;
    ans %= mod;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> mod;
  fac[0] = 1;
  ifac[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = 1LL * fac[i - 1] * i % mod;
    ifac[i] = power(fac[i], mod - 2);
  }
  int ans = 0;
  int lim = 2 * sqrt(n) + 1;
  for (a1 = max(1, n - lim); a1 <= n; a1++) {
    ans += yo(0, 0, n + 1 - a1);
    ans %= mod;
  }
  ans %= mod;
  cout << ans << '\n';
  return 0;
}

```