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>
struct R {
  int mod;
  double inv;
  R(int m) : mod(m), inv(1. / m) {}
  uint64_t operator()(uint64_t x) {
    x -= uint64_t(x * inv + .5) * mod;
    return x < mod ? x : x + mod;
  }
} A(2);
int mod;
inline int inc(int x, int y) { return x += y - mod, x += x >> 31 & mod; }
inline int dec(int x, int y) { return x -= y, x += x >> 31 & mod; }
inline int mul(int x, int y) { return A(uint64_t(x) * y); }
int qpow(int a, int b) {
  int c = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) c = mul(a, c);
  return c;
}
const int maxn = 205;
int n, fac[maxn], fic[maxn], low[maxn], dp[maxn][maxn][maxn];
inline void upd(int &x, int y) { x = inc(x, y); }
void init() {
  fac[0] = 1;
  for (int i = int(1); i <= int(n); i++) fac[i] = mul(fac[i - 1], i);
  fic[n] = qpow(fac[n], mod - 2);
  for (int i = int(n); i >= int(1); i--) fic[i - 1] = mul(fic[i], i);
}
int solve(int m) {
  memset(low, 0, sizeof(low));
  int s = 0;
  for (int i = int(1); i <= int(n - m); i++) {
    s += (low[m + i] = i + 1);
  }
  if (s > m) return 0;
  memset(dp, 0, sizeof(dp));
  for (int i = int(1); i <= int(m); i++) {
    dp[0][i][0] = fic[i];
  }
  for (int i = int(0); i <= int(n - m); i++)
    for (int j = int(0); j <= int(n); j++)
      for (int k = int(0); k <= int(m); k++)
        if (k + n - j <= m) {
          for (int l = int(0); l <= int(n - j); l++)
            if (i + 1 >= low[j + l])
              upd(dp[i + 1][j + l][k + n - j], mul(dp[i][j][k], fic[l]));
        }
  int x = 0;
  for (int i = int(0); i <= int(m); i++) {
    upd(x, dp[n - m + 1][n][i]);
  }
  return mul(x, fac[n]);
}
int main() {
  scanf("%d %d", &n, &mod), A = R(mod);
  init();
  int s = 1;
  for (int m = int(n / 2 + 1); m <= int(n); m++) {
    s = inc(s, solve(m));
  }
  printf("%d\n", s);
  return 0;
}
