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;
  const int INF = 0x3f3f3f3f;
  int n, MOD;
  namespace combination {
  const int MAXN = 200000;
  int fact[MAXN + 1], ifact[MAXN + 1];
  int quick(int k1, int k2) {
    int k3 = 1;
    for (; k2; k2 >>= 1, k1 = 1LL * k1 * k1 % MOD)
      if (k2 & 1) k3 = 1LL * k3 * k1 % MOD;
    return k3;
  }
  int inv(int A) { return quick(A, MOD - 2); }
  void genmath() {
    fact[0] = 1;
    for (int i = 1; i <= MAXN; ++i) fact[i] = 1ll * fact[i - 1] * i % MOD;
    ifact[MAXN] = inv(fact[MAXN]);
    for (int i = MAXN - 1; i >= 0; --i)
      ifact[i] = 1ll * ifact[i + 1] * (i + 1) % MOD;
  }
  int comb(int A, int B) {
    if (A < B) return 0;
    return 1ll * fact[A] * ifact[B] % MOD * ifact[A - B] % MOD;
  }
  void add(int &A, int B) {
    A += B;
    if (A >= MOD) A -= MOD;
  }
  void sub(int &A, int B) {
    A += MOD - B;
    if (A >= MOD) A -= MOD;
  }
  }  // namespace combination
  using namespace combination;
  unordered_map<int, int> dp[202];
  int main() {
    cin >> n >> MOD;
    genmath();
    int ans = 0;
    for (int f = 1; f <= n + 1; ++f) {
      for (int i = 1; i <= n; ++i) dp[i].clear();
      for (int i = f * n; i <= f + (n - 1) * (n + 1); ++i) {
        for (int j = max(f, (i + f - 1) / f); j <= n + 1; ++j) {
          dp[n][i * 202 + j] = 1;
        }
      }
      for (int i = n; i >= 2; --i) {
        for (auto ite = dp[i].begin(); ite != dp[i].end(); ite++) {
          int j, k;
          j = ite->first / 202;
          k = ite->first % 202;
          int val = ite->second;
          for (int prej = i - 1; prej >= 1; --prej) {
            int s = j - (i - prej) * k;
            if (s < 0) break;
            if (s >= f * prej)
              for (int preval = max(f, (s + f - 1) / f); preval <= k - 1;
                   ++preval) {
                if (s >= f * (prej - 1) + preval) {
                  add(dp[prej][s * 202 + preval],
                      1ll * val * ifact[i - prej] % MOD);
                }
              }
          }
        }
      }
      for (int i = 1; i <= n; ++i) {
        add(ans, 1ll * dp[i][i * f * 202 + f] * ifact[i] % MOD * fact[n] % MOD);
      }
    }
    cout << ans << endl;
    return 0;
  }
  