#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &f) {
  f = 0;
  T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') {
      fu = -1;
    }
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + 48);
  else
    print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
  print(x);
  putchar(t);
}
int md;
inline int add(int x, int y) {
  if (x + y >= md) return x + y - md;
  return x + y;
}
inline void addx(int &x, int y) {
  x += y;
  if (x >= md) x -= md;
}
inline int sub(int x, int y) {
  if (x < y) return x - y + md;
  return x - y;
}
inline void subx(int &x, int y) {
  x -= y;
  if (x < 0) x += md;
}
inline int mul(int x, int y) { return 1ull * x * y % md; }
inline int fpow(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    y >>= 1;
    x = mul(x, x);
  }
  return ans;
}
const int N = 205;
int f[N][N][N], fac[N], inv[N];
int n, ans;
bool check(int fir) {
  int sum = fir, now = fir;
  for (int i = 2; i <= n && now <= n + 1; i++) {
    while (fir * now < sum + now && now <= n + 1) ++now;
    sum += now;
  }
  return now <= n + 1;
}
int main() {
  read(n);
  read(md);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = mul(fac[i - 1], i);
  inv[n] = fpow(fac[n], md - 2);
  for (int i = n; i >= 1; i--) inv[i - 1] = mul(inv[i], i);
  for (int mn = 1; mn <= n + 1; mn++) {
    if (!check(mn)) continue;
    memset(f[mn - 1], 0, sizeof(f[mn - 1]));
    f[mn - 1][0][0] = 1;
    for (int i = mn; i <= n + 1; i++) {
      memset(f[i], 0, sizeof(f[i]));
      for (int j = 0; j <= i - 1; j++) {
        for (int k = 0; k <= n - (i - mn) * (n - j); k++) {
          if (!f[i - 1][j][k]) continue;
          int low = (i == mn || i == n + 1);
          for (int t = low;
               t <= (i == mn ? i - j : min(min(n, i) - j, (n - k) / (i - mn)));
               t++) {
            if (i * mn < j * mn + k + i * t) break;
            f[i][j + t][k + (i - mn) * t] =
                add(f[i][j + t][k + (i - mn) * t], mul(f[i - 1][j][k], inv[t]));
          }
        }
      }
    }
    for (int i = 0; i <= n; i++) ans = add(ans, f[n + 1][n][i]);
  }
  print(add(mul(ans, fac[n]), 1), '\n');
  return 0;
}
