#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
const int xn = 50 + 10;
const int xm = -20;
const int sq = 320;
const int inf = 1e9 + 10;
const long long INF = 1e18 + 10;
const int mod = 1e9 + 7;
const int base = 257;
long long n, k, dp[xn][xn][2], fact[xn], ifact[xn];
vector<pair<pair<int, int>, pair<int, int> > > vec[xn + xn][2];
long long C(long long r, long long n) {
  if (r < 0 || n - r < 0) return 0;
  return fact[n] * ifact[r] % mod * ifact[n - r] % mod;
}
int main() {
  fact[0] = 1;
  for (long long i = 1; i < xn; ++i) fact[i] = fact[i - 1] * i % mod;
  ifact[xn - 1] = power(fact[xn - 1], mod - 2, mod);
  for (long long i = xn - 2; i >= 0; --i)
    ifact[i] = ifact[i + 1] * (i + 1) % mod;
  cin >> n >> k;
  for (int i = 0; i < xn; ++i) {
    for (int f1 = 0; f1 < 2; ++f1) {
      for (int j = 0; j < xn; ++j) {
        for (int f2 = 0; f2 < 2; ++f2) {
          if (i - f1 < 0 || j - f2 < 0) continue;
          pair<int, int> x = {i + j + 1 - min(f1, f2), i + j};
          x.second = bool(x.first > x.second);
          vec[x.first][x.second].push_back({{i, f1}, {j, f2}});
        }
      }
    }
  }
  dp[1][0][0] = 1;
  for (long long s = 2; s <= n; ++s) {
    for (long long t = 1; t <= k; ++t) {
      dp[s][t][0] = dp[s - 1][t][1] * s % mod;
      dp[s][t][1] = dp[s - 1][t - 1][0] * s % mod;
      for (int f = 0; f < 2; ++f) {
        for (pair<pair<int, int>, pair<int, int> > x : vec[t][f]) {
          for (long long i = 1; i < s - 1; ++i) {
            long long res = dp[i][x.first.first][x.first.second] *
                            dp[s - i - 1][x.second.first][x.second.second] %
                            mod;
            res = res * C(i - 1, s - 2) % mod * s % mod;
            dp[s][t][f] = (dp[s][t][f] + res) % mod;
          }
        }
      }
    }
  }
  cout << (dp[n][k][0] + dp[n][k][1]) * power(n, mod - 2, mod) % mod << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

Ksenia has her winter exams. Today she is learning combinatorics. Here's one of the problems she needs to learn to solve.

How many distinct trees are there consisting of n vertices, each with the following properties:

  * the tree is marked, that is, the vertices of the tree are numbered from 1 to n; 
  * each vertex of the tree is connected with at most three other vertices, and at the same moment the vertex with number 1 is connected with at most two other vertices; 
  * the size of the tree's maximum matching equals k. 



Two trees are considered distinct if there are such two vertices u and v, that in one tree they are connected by an edge and in the other tree they are not.

Help Ksenia solve the problem for the given n and k. As the answer to the problem can be very huge you should output it modulo 1000000007 (109 + 7).

Input

The first line contains two integers n, k (1 ≤ n, k ≤ 50).

Output

Print a single integer — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

1 1


Output

0


Input

2 1


Output

1


Input

3 1


Output

3


Input

4 2


Output

12

Note

If you aren't familiar with matchings, please, read the following link: http://en.wikipedia.org/wiki/Matching_(graph_theory).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
const int xn = 50 + 10;
const int xm = -20;
const int sq = 320;
const int inf = 1e9 + 10;
const long long INF = 1e18 + 10;
const int mod = 1e9 + 7;
const int base = 257;
long long n, k, dp[xn][xn][2], fact[xn], ifact[xn];
vector<pair<pair<int, int>, pair<int, int> > > vec[xn + xn][2];
long long C(long long r, long long n) {
  if (r < 0 || n - r < 0) return 0;
  return fact[n] * ifact[r] % mod * ifact[n - r] % mod;
}
int main() {
  fact[0] = 1;
  for (long long i = 1; i < xn; ++i) fact[i] = fact[i - 1] * i % mod;
  ifact[xn - 1] = power(fact[xn - 1], mod - 2, mod);
  for (long long i = xn - 2; i >= 0; --i)
    ifact[i] = ifact[i + 1] * (i + 1) % mod;
  cin >> n >> k;
  for (int i = 0; i < xn; ++i) {
    for (int f1 = 0; f1 < 2; ++f1) {
      for (int j = 0; j < xn; ++j) {
        for (int f2 = 0; f2 < 2; ++f2) {
          if (i - f1 < 0 || j - f2 < 0) continue;
          pair<int, int> x = {i + j + 1 - min(f1, f2), i + j};
          x.second = bool(x.first > x.second);
          vec[x.first][x.second].push_back({{i, f1}, {j, f2}});
        }
      }
    }
  }
  dp[1][0][0] = 1;
  for (long long s = 2; s <= n; ++s) {
    for (long long t = 1; t <= k; ++t) {
      dp[s][t][0] = dp[s - 1][t][1] * s % mod;
      dp[s][t][1] = dp[s - 1][t - 1][0] * s % mod;
      for (int f = 0; f < 2; ++f) {
        for (pair<pair<int, int>, pair<int, int> > x : vec[t][f]) {
          for (long long i = 1; i < s - 1; ++i) {
            long long res = dp[i][x.first.first][x.first.second] *
                            dp[s - i - 1][x.second.first][x.second.second] %
                            mod;
            res = res * C(i - 1, s - 2) % mod * s % mod;
            dp[s][t][f] = (dp[s][t][f] + res) % mod;
          }
        }
      }
    }
  }
  cout << (dp[n][k][0] + dp[n][k][1]) * power(n, mod - 2, mod) % mod << '\n';
  return 0;
}

```