#include <bits/stdc++.h>
using namespace std;
void per() { cerr << endl; }
template <typename Head, typename... Tail>
void per(Head H, Tail... T) {
  cerr << H << ' ';
  per(T...);
}
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
template <class U, class V>
ostream& operator<<(ostream& out, const pair<U, V>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <class U, class V>
istream& operator>>(istream& in, pair<U, V>& a) {
  return in >> a.first >> a.second;
}
template <typename W,
          typename T = typename enable_if<!is_same<W, string>::value,
                                          typename W::value_type>::type>
ostream& operator<<(ostream& out, const W& v) {
  out << "{ ";
  for (const auto& first : v) out << first << ", ";
  return out << '}';
}
template <class T>
void readArr(T from, T to) {
  for (auto i = from; i != to; ++i) cin >> *i;
}
mt19937 mrand(1337);
unsigned int myRand32() { return mrand() & (unsigned int)(-1); }
unsigned long long myRand64() {
  return ((unsigned long long)myRand32() << 32) ^ myRand32();
}
int mod = 1000000007;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void dec(int& a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mult(int a, int b) { return a * (long long)b % mod; }
int bp(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mult(res, a);
    a = mult(a, a);
    b >>= 1;
  }
  return res;
}
const int N = 207;
int C[N][N];
int dp[N][N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n >> mod;
  C[0][0] = 1;
  for (int i = 1; i < N; ++i) {
    C[i][0] = 1, C[i][i] = 1;
    for (int j = 1; j < i; ++j) {
      C[i][j] = C[i - 1][j];
      add(C[i][j], C[i - 1][j - 1]);
    }
  }
  int ans = 1;
  for (int min_element = 1; min_element <= n + 1; ++min_element) {
    for (int i = min_element; i <= n + 1; ++i) {
      int can_take = min_element / (i - min_element + 1);
      for (int take = max(0, n - can_take); take <= n; ++take) {
        for (int excess = 0; excess <= min_element; ++excess) {
          dp[i][take][excess] = 0;
        }
      }
    }
    for (int take = 1; take <= n; ++take) {
      add(dp[min_element][take][0], C[n][take]);
    }
    for (int i = min_element + 1; i <= n + 1; ++i) {
      int can_take = min_element / (i - min_element);
      for (int take = max(1, n - can_take); take < n; ++take) {
        for (int excess = 0; excess <= min_element; ++excess) {
          if (!dp[i - 1][take][excess]) continue;
          if (min_element * (i - 1) < min_element * take + excess) continue;
          for (int want = 0; want <= can_take &&
                             excess + (i - min_element) * want <= min_element;
               ++want) {
            int new_excess = excess + (i - min_element) * want;
            if (min_element * i < min_element * (take + want) + new_excess)
              break;
            add(dp[i][take + want][new_excess],
                mult(dp[i - 1][take][excess], C[n - take][want]));
          }
        }
      }
    }
    for (int excess = 0; excess <= min_element; ++excess) {
      add(ans, dp[n + 1][n][excess]);
    }
  }
  cout << ans;
}

### 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>
using namespace std;
void per() { cerr << endl; }
template <typename Head, typename... Tail>
void per(Head H, Tail... T) {
  cerr << H << ' ';
  per(T...);
}
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
template <class U, class V>
ostream& operator<<(ostream& out, const pair<U, V>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <class U, class V>
istream& operator>>(istream& in, pair<U, V>& a) {
  return in >> a.first >> a.second;
}
template <typename W,
          typename T = typename enable_if<!is_same<W, string>::value,
                                          typename W::value_type>::type>
ostream& operator<<(ostream& out, const W& v) {
  out << "{ ";
  for (const auto& first : v) out << first << ", ";
  return out << '}';
}
template <class T>
void readArr(T from, T to) {
  for (auto i = from; i != to; ++i) cin >> *i;
}
mt19937 mrand(1337);
unsigned int myRand32() { return mrand() & (unsigned int)(-1); }
unsigned long long myRand64() {
  return ((unsigned long long)myRand32() << 32) ^ myRand32();
}
int mod = 1000000007;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void dec(int& a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mult(int a, int b) { return a * (long long)b % mod; }
int bp(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mult(res, a);
    a = mult(a, a);
    b >>= 1;
  }
  return res;
}
const int N = 207;
int C[N][N];
int dp[N][N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n >> mod;
  C[0][0] = 1;
  for (int i = 1; i < N; ++i) {
    C[i][0] = 1, C[i][i] = 1;
    for (int j = 1; j < i; ++j) {
      C[i][j] = C[i - 1][j];
      add(C[i][j], C[i - 1][j - 1]);
    }
  }
  int ans = 1;
  for (int min_element = 1; min_element <= n + 1; ++min_element) {
    for (int i = min_element; i <= n + 1; ++i) {
      int can_take = min_element / (i - min_element + 1);
      for (int take = max(0, n - can_take); take <= n; ++take) {
        for (int excess = 0; excess <= min_element; ++excess) {
          dp[i][take][excess] = 0;
        }
      }
    }
    for (int take = 1; take <= n; ++take) {
      add(dp[min_element][take][0], C[n][take]);
    }
    for (int i = min_element + 1; i <= n + 1; ++i) {
      int can_take = min_element / (i - min_element);
      for (int take = max(1, n - can_take); take < n; ++take) {
        for (int excess = 0; excess <= min_element; ++excess) {
          if (!dp[i - 1][take][excess]) continue;
          if (min_element * (i - 1) < min_element * take + excess) continue;
          for (int want = 0; want <= can_take &&
                             excess + (i - min_element) * want <= min_element;
               ++want) {
            int new_excess = excess + (i - min_element) * want;
            if (min_element * i < min_element * (take + want) + new_excess)
              break;
            add(dp[i][take + want][new_excess],
                mult(dp[i - 1][take][excess], C[n - take][want]));
          }
        }
      }
    }
    for (int excess = 0; excess <= min_element; ++excess) {
      add(ans, dp[n + 1][n][excess]);
    }
  }
  cout << ans;
}

```