#include <bits/stdc++.h>
template <typename _Tp>
void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
const int N = 205;
int n, mod;
template <typename _Tp1, typename _Tp2>
inline void add(_Tp1 &a, _Tp2 b) {
  a = a + b >= mod ? a + b - mod : a + b;
}
template <typename _Tp1, typename _Tp2>
inline void sub(_Tp1 &a, _Tp2 b) {
  a = a - b < 0 ? a - b + mod : a - b;
}
long long fac[N], inv[N], ifac[N];
void init() {
  fac[0] = fac[1] = inv[0] = inv[1] = ifac[0] = ifac[1] = 1;
  for (int i = 2; i < N; ++i)
    fac[i] = fac[i - 1] * i % mod,
    inv[i] = inv[mod % i] * (mod - mod / i) % mod,
    ifac[i] = ifac[i - 1] * inv[i] % mod;
}
int dp[35][N][N];
int solve(int a1) {
  memset(dp, 0, sizeof(dp)), dp[n + 2 - a1][0][0] = 1;
  for (int i = n + 1 - a1, t = 0; i >= 1; --i, ++t) {
    for (int j = 0; j <= n; ++j) {
      for (int k = 0; k <= n; ++k)
        if (dp[i + 1][j][k]) {
          for (int c = 0; c * i + k <= a1 && c + j <= n; ++c) {
            add(dp[i][j + c][k + c * i], 1LL * dp[i + 1][j][k] * ifac[c] % mod);
          }
        }
    }
    if (i > 1)
      for (int j = 0; j <= t; ++j) memset(dp[i][j], 0, (n + 3) << 2);
  }
  int ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= a1; ++j)
      add(ans, 1LL * dp[1][i][j] * ifac[n - i] % mod);
  return fac[n] * ans % mod;
}
int main() {
  read(n, mod);
  init();
  int ans = 1;
  for (int x = std::max(1, n - 30); x <= n; ++x) add(ans, solve(x));
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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>
template <typename _Tp>
void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
const int N = 205;
int n, mod;
template <typename _Tp1, typename _Tp2>
inline void add(_Tp1 &a, _Tp2 b) {
  a = a + b >= mod ? a + b - mod : a + b;
}
template <typename _Tp1, typename _Tp2>
inline void sub(_Tp1 &a, _Tp2 b) {
  a = a - b < 0 ? a - b + mod : a - b;
}
long long fac[N], inv[N], ifac[N];
void init() {
  fac[0] = fac[1] = inv[0] = inv[1] = ifac[0] = ifac[1] = 1;
  for (int i = 2; i < N; ++i)
    fac[i] = fac[i - 1] * i % mod,
    inv[i] = inv[mod % i] * (mod - mod / i) % mod,
    ifac[i] = ifac[i - 1] * inv[i] % mod;
}
int dp[35][N][N];
int solve(int a1) {
  memset(dp, 0, sizeof(dp)), dp[n + 2 - a1][0][0] = 1;
  for (int i = n + 1 - a1, t = 0; i >= 1; --i, ++t) {
    for (int j = 0; j <= n; ++j) {
      for (int k = 0; k <= n; ++k)
        if (dp[i + 1][j][k]) {
          for (int c = 0; c * i + k <= a1 && c + j <= n; ++c) {
            add(dp[i][j + c][k + c * i], 1LL * dp[i + 1][j][k] * ifac[c] % mod);
          }
        }
    }
    if (i > 1)
      for (int j = 0; j <= t; ++j) memset(dp[i][j], 0, (n + 3) << 2);
  }
  int ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= a1; ++j)
      add(ans, 1LL * dp[1][i][j] * ifac[n - i] % mod);
  return fac[n] * ans % mod;
}
int main() {
  read(n, mod);
  init();
  int ans = 1;
  for (int x = std::max(1, n - 30); x <= n; ++x) add(ans, solve(x));
  printf("%d\n", ans);
  return 0;
}

```