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.
10 999999937
 4 998244353
 198 549999959
 22 100757149
 2 549999959
 103 267850993
 1 550000001
 192 999999937
 6 998244353
 194 100000007
 162 999999937
 3 549999959
 140 550000001
 106 100757149
 120 267850993
 170 100000007
 93 267850993
 196 100000007
 191 100757149
 20 549999959
 9 999999937
 1 100000007
 130 100757149
 197 100000007
 192 998244353
 160 100757149
 193 998244353
 110 999999937
 190 999999937
 10 267850993
 7 100000007
 100 267850993
 82 267850993
 8 549999959
 200 550000001
 200 998244353
 169 100757149
 195 549999959
 119 999999937
 50 100757149
 196 100757149
 180 100757149
 5 100757149
 28 549999959
 150 998244353
 200 549999959
 188 100757149
 190 267850993
 199 550000001

38324
 32
 545511297
 12864320
 4
 262873012
 2
 960726876
 295
 24964109
 356680016
 11
 464259310
 7881104
 91785690
 28293730
 34145854
 80552289
 22329873
 461862358
 10901
 2
 87253785
 59296946
 216873641
 54532638
 959300190
 165234842
 259139303
 38324
 940
 189712610
 181769218
 3148
 481731577
 180611733
 41487196
 305462979
 810859890
 4745300
 71649802
 40944895
 97
 265377742
 563043130
 178617235
 15934891
 38309988
 262511266

#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;
}
