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;
  char buf[100000], *p1(buf), *p2(buf);
  inline int read() {
    int x = 0, f = 1;
    char ch =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++);
    while (!isdigit(ch)) {
      if (ch == '-') f = 0;
      ch = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
                ? EOF
                : *p1++);
    }
    while (isdigit(ch)) {
      x = (x << 1) + (x << 3) + (ch ^ 48);
      ch = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
                ? EOF
                : *p1++);
    }
    return f ? x : -x;
  }
  using ll = long long;
  const int N = 2e2 + 10;
  int n, mod;
  int fc[N], fv[N];
  inline void chk(int &x) {
    x -= mod;
    x += x >> 31 & mod;
  }
  inline int mll(int x, int y) { return (ll)x * y % mod; }
  inline int ksm(int x, int y) {
    int ret = 1;
    for (; y; y >>= 1, x = mll(x, x))
      if (y & 1) ret = mll(ret, x);
    return ret;
  }
  inline void input() {
    n = read(), mod = read();
    return;
  }
  inline void pref(const int lim) {
    fc[0] = 1;
    for (int i = 1; i <= lim; i++) fc[i] = mll(fc[i - 1], i);
    fv[lim] = ksm(fc[lim], mod - 2);
    for (int i = lim; i >= 1; i--) fv[i - 1] = mll(fv[i], i);
    return;
  }
  bool vis[N][N][50];
  int dp[N][N][50];
  int frst;
  int calc(int cnt, int sum, int cur, int lim) {
    if (vis[cnt][sum][cur]) return dp[cnt][sum][cur];
    vis[cnt][sum][cur] = true;
    int &ret = dp[cnt][sum][cur];
    ret = 0;
    if (cur <= lim - 1 && cnt < cur) return ret;
    if (cur == lim) {
      if (cnt < n) ret = fv[n - cnt];
      return ret;
    }
    int x = n + 1 - cur - frst;
    for (int j = 0; cnt + j <= n && sum + j * x <= frst; j++)
      chk(ret += mll(calc(cnt + j, sum + j * x, cur + 1, lim), fv[j]));
    return ret;
  }
  inline void work() {
    pref(n);
    int ans = 0;
    for (int lim = 0; lim <= 32; lim++) {
      memset(vis, 0, sizeof(vis));
      frst = n + 1 - lim;
      chk(ans += calc(0, 0, 0, lim));
    }
    cout << mll(ans, fc[n]) << '\n';
    return;
  }
  inline void solve() {
    input();
    work();
    return;
  }
  int main() {
    solve();
    return 0;
  }
  