#include <bits/stdc++.h>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 205;
int fac[N], ifac[N];
int n, mod, ans, a1;
int dp[N][N][45], vis[N][N][45];
inline int qpow(int a, int b = mod - 2) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return res;
}
int dfs(int i, int sum, int k) {
  if (i == n) return 1;
  if (!k) return ifac[n - i];
  if (vis[i][sum][k] == a1) return dp[i][sum][k];
  vis[i][sum][k] = a1;
  int &res = dp[i][sum][k];
  res = 0;
  int lim = (a1 - sum) / k;
  for (int cnt = 0; cnt <= lim; cnt++) {
    if (k > 1 && i + cnt < n + 1 - a1 - (k - 1)) continue;
    res = (res + 1ll * dfs(i + cnt, sum + cnt * k, k - 1) * ifac[cnt]) % mod;
  }
  return res;
}
int main() {
  n = read(), mod = read();
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= N - 1; i++) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
    ifac[i] = qpow(fac[i]);
  }
  int lim = 2 * sqrt(n) + 1;
  for (a1 = max(1, n - lim); a1 <= n; a1++) {
    ans = (ans + dfs(0, 0, n + 1 - a1)) % mod;
  }
  print(1ll * fac[n] * ans % 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>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 205;
int fac[N], ifac[N];
int n, mod, ans, a1;
int dp[N][N][45], vis[N][N][45];
inline int qpow(int a, int b = mod - 2) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return res;
}
int dfs(int i, int sum, int k) {
  if (i == n) return 1;
  if (!k) return ifac[n - i];
  if (vis[i][sum][k] == a1) return dp[i][sum][k];
  vis[i][sum][k] = a1;
  int &res = dp[i][sum][k];
  res = 0;
  int lim = (a1 - sum) / k;
  for (int cnt = 0; cnt <= lim; cnt++) {
    if (k > 1 && i + cnt < n + 1 - a1 - (k - 1)) continue;
    res = (res + 1ll * dfs(i + cnt, sum + cnt * k, k - 1) * ifac[cnt]) % mod;
  }
  return res;
}
int main() {
  n = read(), mod = read();
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= N - 1; i++) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
    ifac[i] = qpow(fac[i]);
  }
  int lim = 2 * sqrt(n) + 1;
  for (a1 = max(1, n - lim); a1 <= n; a1++) {
    ans = (ans + dfs(0, 0, n + 1 - a1)) % mod;
  }
  print(1ll * fac[n] * ans % mod, '\n');
  return 0;
}

```