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;
long long read() {
  long long x;
  scanf("%lld", &x);
  return x;
}
int m;
int add(int a, int b) { return a + b < (m) ? a + b : a + b - (m); }
void dadd(int &a, int b) { a = add(a, b); }
int sub(int a, int b) { return a >= b ? a - b : a - b + (m); }
void dsub(int &a, int b) { a = sub(a, b); }
int mul(int a, int b) { return (long long)a * b % (m); }
void dmul(int &a, int b) { a = mul(a, b); }
int ksm(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) dmul(ans, a);
    b >>= 1;
    dmul(a, a);
  }
  return ans;
}
int inv(int a) { return ksm(a, (m)-2); }
int ceil(int a, int b) { return (a - 1) / b + 1; }
const int N = 202;
const int K = 25;
int n;
int fr[N];
int ifr[N];
void init() {
  for (int i = (int)(0); i <= (int)(n); i++) fr[i] = i ? mul(fr[i - 1], i) : 1;
  for (int i = (int)(0); i <= (int)(n); i++) ifr[i] = inv(fr[i]);
}
bool done[K][N][K * (N + 1)];
int dp[K][N][K * (N + 1)];
int dfs(int a1, int x, int mn, int s) {
  if (x == n) return fr[n];
  if (mn > n + 1) return 0;
  if (a1 * (n + 1) < s + mn * (n - x)) return 0;
  bool &ok = done[n + 1 - mn][x][x * (n + 1) - s];
  int &aans = dp[n + 1 - mn][x][x * (n + 1) - s];
  if (ok) return aans;
  int ans = 0;
  ok = true;
  if (x == 0) {
    int t = s;
    for (int y = (int)(x + 1); y <= (int)(n); y++) {
      t += a1;
      if (t > a1 * a1) break;
      dadd(ans, mul(ifr[y - x], dfs(a1, y, a1 + 1, t)));
    }
  } else {
    for (int v = (int)(mn); v <= (int)(n + 1); v++) {
      int t = s;
      for (int y = (int)(x + 1); y <= (int)(n); y++) {
        t += v;
        if (t > a1 * v) break;
        dadd(ans, mul(ifr[y - x], dfs(a1, y, v + 1, t)));
      }
    }
  }
  aans = ans;
  return ans;
}
int calc(int a1) {
  memset(done, false, sizeof(done));
  return dfs(a1, 0, a1, 0);
}
bool check(int a1) {
  int ax = a1;
  int sum = a1;
  for (int k = (int)(2); k <= (int)(n); k++) {
    while (a1 * ax < sum + ax && ax <= n + 1) ax++;
    sum += ax;
  }
  return ax <= n + 1;
}
int main() {
  n = read();
  m = read();
  init();
  int ans = 0;
  for (int j = (int)(1); j <= (int)(n + 1); j++)
    if (check(j)) dadd(ans, calc(j));
  printf("%d\n", ans);
  return 0;
}
