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, typename S>
ostream& operator<<(ostream& out, const pair<T, S>& p) {
  out << "(" << p.first << " " << p.second << ")";
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  out << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    out << v[i];
    if (i != (int)v.size() - 1) out << ", ";
  }
  out << "]";
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<vector<T> >& m) {
  for (int i = 0; i < m.size(); ++i) {
    out << m[i] << "\n";
  }
  return out;
}
template <typename H1>
ostream& show(ostream& out, const char* label, H1&& value) {
  return out << label << " = " << value << '\n';
}
template <typename H1, typename... T>
ostream& show(ostream& out, const char* label, H1&& value, T&&... rest) {
  const char* first_comma = strchr(label, ',');
  const char* left_parenthesis = strchr(label, '(');
  if (left_parenthesis != nullptr && left_parenthesis < first_comma) {
    const char* right_parenthesis = strchr(left_parenthesis, ')');
    assert(right_parenthesis != nullptr);
    const char* pcomma = strchr(right_parenthesis, ',');
    return show(out.write(label, pcomma - label) << " = " << value << ',',
                pcomma + 1, rest...);
  }
  return show(out.write(label, first_comma - label) << " = " << value << ',',
              first_comma + 1, rest...);
}
const int MAX_N = 200 + 2;
long long fact[MAX_N];
long long fact_inv[MAX_N];
int m, n, M;
long long dp[MAX_N][MAX_N][MAX_N];
long long power(long long a, int pow) {
  if (pow == 0) {
    return 1;
  }
  long long half = power(a, pow / 2);
  return ((half * half % M) * (pow & 1 ? a : 1)) % M;
}
void preprocess() {
  fact[0] = 1;
  fact_inv[0] = 1;
  for (int i = 1; i < MAX_N; ++i) {
    fact[i] = fact[i - 1] * i % M;
    fact_inv[i] = power(fact[i], M - 2);
  }
}
long long solve(int i, int k, int s) {
  if (k > n + 1) {
    return 0;
  }
  if (k == n + 1) {
    return fact_inv[n - i + 1];
  }
  if (i == n) {
    return 1;
  }
  long long& ans = dp[i][k][s];
  if (ans != -1) {
    return ans;
  }
  ans = 0;
  const int min_t = max(0, (k + 1 - m) * (n - i) + s - 2 * m + n + 1);
  const int max_t = min(n - i, (k * m - s - m * (i - 1)) / k);
  for (int t = min_t; t <= max_t; ++t) {
    ans += fact_inv[t] * solve(i + t, k + 1, s + (k - m) * t);
    ans %= M;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> M;
  preprocess();
  long long ans = 0;
  for (m = (n + 2) / 2; m <= n; ++m) {
    memset(dp, -1, sizeof dp);
    const int min_t = max(1, 2 * (n - m));
    const int max_t = min(n - 1, m);
    for (int t = min_t; t <= max_t; ++t) {
      ans += fact_inv[t] * solve(1 + t, m + 1, 0);
      ans %= M;
    }
  }
  cout << (ans * fact[n] + 2) % M;
}
