description: 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.
solution: #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;
  }
  