#include <bits/stdc++.h>
using namespace std;
const int N = 59;
const int Mod = 1e9 + 7;
int dp[N][N][2], c[N][N];
long long ni2;
long long Pow(long long a, long long b, long long c) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % c;
    a = a * a % c;
    b >>= 1;
  }
  return ret;
}
int dfs(int n, int k, int t) {
  if (n == 0) return (k == 0) && (t == 1);
  if (n == 1) return (k == 0) && (t == 0);
  int &ret = dp[n][k][t];
  if (ret + 1) return ret;
  ret = 0;
  for (int n1 = 0; n1 < n; n1++) {
    int n2 = n - n1 - 1;
    if (n1 > n2) continue;
    for (int k1 = 0; k1 + t <= k; k1++) {
      int k2 = k - k1 - t;
      for (int t1 = 0; t1 < 2; t1++)
        for (int t2 = 0; t2 < 2; t2++) {
          if (t == 0) {
            if (t1 == 0 || t2 == 0) continue;
            long long T = 1ll * dfs(n1, k1, t1) * dfs(n2, k2, t2) % Mod;
            T = T * c[n][n1] % Mod * c[n - n1][n2] % Mod;
            if (n1 == n2) T = T * ni2 % Mod;
            ret = (ret + T) % Mod;
          } else {
            if (t1 && t2) continue;
            long long T = 1ll * dfs(n1, k1, t1) * dfs(n2, k2, t2) % Mod;
            T = T * c[n][n1] % Mod * c[n - n1][n2] % Mod;
            if (n1 == n2) T = T * ni2 % Mod;
            ret = (ret + T) % Mod;
          }
        }
    }
  }
  return ret;
}
int main() {
  ni2 = Pow(2, Mod - 2, Mod);
  memset(dp, -1, sizeof(dp));
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < N; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % Mod;
  }
  long long ans = dfs(n, k, 0) + dfs(n, k, 1) % Mod;
  ans = ans * Pow(n, Mod - 2, Mod) % Mod;
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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;
const int N = 59;
const int Mod = 1e9 + 7;
int dp[N][N][2], c[N][N];
long long ni2;
long long Pow(long long a, long long b, long long c) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % c;
    a = a * a % c;
    b >>= 1;
  }
  return ret;
}
int dfs(int n, int k, int t) {
  if (n == 0) return (k == 0) && (t == 1);
  if (n == 1) return (k == 0) && (t == 0);
  int &ret = dp[n][k][t];
  if (ret + 1) return ret;
  ret = 0;
  for (int n1 = 0; n1 < n; n1++) {
    int n2 = n - n1 - 1;
    if (n1 > n2) continue;
    for (int k1 = 0; k1 + t <= k; k1++) {
      int k2 = k - k1 - t;
      for (int t1 = 0; t1 < 2; t1++)
        for (int t2 = 0; t2 < 2; t2++) {
          if (t == 0) {
            if (t1 == 0 || t2 == 0) continue;
            long long T = 1ll * dfs(n1, k1, t1) * dfs(n2, k2, t2) % Mod;
            T = T * c[n][n1] % Mod * c[n - n1][n2] % Mod;
            if (n1 == n2) T = T * ni2 % Mod;
            ret = (ret + T) % Mod;
          } else {
            if (t1 && t2) continue;
            long long T = 1ll * dfs(n1, k1, t1) * dfs(n2, k2, t2) % Mod;
            T = T * c[n][n1] % Mod * c[n - n1][n2] % Mod;
            if (n1 == n2) T = T * ni2 % Mod;
            ret = (ret + T) % Mod;
          }
        }
    }
  }
  return ret;
}
int main() {
  ni2 = Pow(2, Mod - 2, Mod);
  memset(dp, -1, sizeof(dp));
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < N; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % Mod;
  }
  long long ans = dfs(n, k, 0) + dfs(n, k, 1) % Mod;
  ans = ans * Pow(n, Mod - 2, Mod) % Mod;
  cout << ans << endl;
  return 0;
}

```