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;
constexpr int MaxN = 200 + 5, MaxSqrt = 20 + 5;
constexpr long long Infinity = (1LL << 62);
inline int QuickPow(const int val, const int lev, const int Mod) {
  if (!lev) return 1;
  if (lev & 1) return ((long long)QuickPow(val, lev ^ 1, Mod) * val) % Mod;
  const int Half = QuickPow(val, lev >> 1, Mod);
  return ((long long)Half * Half) % Mod;
}
inline int Inverse(const int val, const int Mod) {
  return QuickPow(val, Mod - 2, Mod);
}
int Fac[MaxN], Inv[MaxN], dp[MaxN][MaxSqrt][MaxN];
bitset<MaxN> Bitset[MaxN][MaxSqrt];
inline int Dynamic_Programming(const int n, const int a, const int i,
                               const int x, const int Sum, const int Mod) {
  if (Bitset[i][x][Sum]) return dp[i][x][Sum];
  Bitset[i][x][Sum] = true;
  long long Record = 0;
  if (!x)
    Record = (((i <= a) && (!Sum)) ? ((long long)Fac[n] * Inv[i]) : 0);
  else {
    const int Lef = ((i <= a) ? 0 : ((i + 1) - a)), Rig = ((n + 1) - a);
    if ((Lef <= x) && (x <= Rig))
      for (register int j = (i); j >= (1); --j) {
        if (Sum < ((i - j) * x)) break;
        if ((Record +=
             ((long long)Inv[i - j] *
              Dynamic_Programming(n, a, j, x - 1, Sum - ((i - j) * x), Mod))) &
            Infinity)
          Record %= Mod;
      }
  }
  return (dp[i][x][Sum] = (Record % Mod));
}
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  int n, Mod;
  cin >> n >> Mod, Fac[0] = 1;
  for (register int i = (1); i <= (n); ++i)
    Fac[i] = ((long long)Fac[i - 1] * i) % Mod;
  Inv[n] = Inverse(Fac[n], Mod);
  for (register int i = (n); i >= (1); --i)
    Inv[i - 1] = ((long long)Inv[i] * i) % Mod;
  const int Limit = max(1, (int)((n - sqrt(n << 1)) - 2));
  long long Answer = 0;
  for (register int a = (Limit); a <= (n + 1); ++a) {
    memset(dp, 0, sizeof(dp));
    memset(Bitset, 0, sizeof(Bitset));
    for (register int i = (0); i <= (a); ++i)
      Answer += Dynamic_Programming(n, a, n, ((n + 1) - a), i, Mod);
  }
  cout << (Answer %= Mod) << endl;
  return 0;
}
