#include <bits/stdc++.h>
using namespace std;
const double pi = acos(0.0) * 2.0;
const double eps = 1e-12;
const int step[8][2] = {{1, 0}, {-1, 0}, {0, 1},  {0, -1},
                        {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <class T>
inline T max1(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline T max1(T a, T b, T c) {
  return max1(max1(a, b), c);
}
template <class T>
inline T max1(T a, T b, T c, T d) {
  return max1(max1(a, b, c), d);
}
template <class T>
inline T max1(T a, T b, T c, T d, T e) {
  return max1(max1(a, b, c, d), e);
}
template <class T>
inline T min1(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline T min1(T a, T b, T c) {
  return min1(min1(a, b), c);
}
template <class T>
inline T min1(T a, T b, T c, T d) {
  return min1(min1(a, b, c), d);
}
template <class T>
inline T min1(T a, T b, T c, T d, T e) {
  return min1(min1(a, b, c, d), e);
}
template <class T>
inline T lowb(T num) {
  return num & (-num);
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / abs1(a) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
inline int bitnum(unsigned int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(unsigned long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitnum(long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitmaxl(unsigned int a) {
  if (a) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(int a) {
  if (a) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(unsigned long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
inline int bitmaxl(long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
long long pow(long long n, long long m, long long mod = 0) {
  if (m < 0) return 0;
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
const int mod = 1000000007;
const int maxn = 55;
long long dp[maxn][maxn][2];
long long c[maxn][maxn];
int n, maxm;
inline void add(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d%d", &n, &maxm);
  memset(dp, 0, sizeof(dp));
  dp[1][0][0] = 1;
  for (int i = 0; i <= n; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      if (c[i][j] >= mod) c[i][j] -= mod;
    }
  }
  for (int i = 2; i <= n; i++)
    for (int j = 0; j <= i - 1; j++) {
      add(dp[i][j][0], dp[i - 1][j][1] * i % mod);
      add(dp[i][j][1], dp[i - 1][j - 1][0] * i % mod);
      for (int k = 1; k <= (i - 1) / 2; k++) {
        int maxt = k / 2;
        for (int t = 0; t <= maxt; t++) {
          long long add0 = dp[k][t][1] * dp[i - k - 1][j - t][1] % mod;
          long long add1 = 0;
          if (t < j) {
            add(add1, dp[k][t][1] * dp[i - k - 1][j - t - 1][0] % mod);
            add(add1, dp[k][t][0] * dp[i - k - 1][j - t - 1][1] % mod);
            add(add1, dp[k][t][0] * dp[i - k - 1][j - t - 1][0] % mod);
          }
          if (k == i - k - 1) {
            add0 *= c[i - 2][k - 1];
            add0 %= mod;
            add1 *= c[i - 2][k - 1];
            add1 %= mod;
          } else {
            add0 *= c[i - 1][k];
            add0 %= mod;
            add1 *= c[i - 1][k];
            add1 %= mod;
          }
          add(dp[i][j][0], add0 * i % mod);
          add(dp[i][j][1], add1 * i % mod);
        }
      }
    }
  cout << (dp[n][maxm][0] + dp[n][maxm][1]) % mod * pow(n, mod - 2, mod) % mod
       << endl;
  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;
const double pi = acos(0.0) * 2.0;
const double eps = 1e-12;
const int step[8][2] = {{1, 0}, {-1, 0}, {0, 1},  {0, -1},
                        {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <class T>
inline T max1(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline T max1(T a, T b, T c) {
  return max1(max1(a, b), c);
}
template <class T>
inline T max1(T a, T b, T c, T d) {
  return max1(max1(a, b, c), d);
}
template <class T>
inline T max1(T a, T b, T c, T d, T e) {
  return max1(max1(a, b, c, d), e);
}
template <class T>
inline T min1(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline T min1(T a, T b, T c) {
  return min1(min1(a, b), c);
}
template <class T>
inline T min1(T a, T b, T c, T d) {
  return min1(min1(a, b, c), d);
}
template <class T>
inline T min1(T a, T b, T c, T d, T e) {
  return min1(min1(a, b, c, d), e);
}
template <class T>
inline T lowb(T num) {
  return num & (-num);
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / abs1(a) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
inline int bitnum(unsigned int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(unsigned long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitnum(long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitmaxl(unsigned int a) {
  if (a) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(int a) {
  if (a) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(unsigned long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
inline int bitmaxl(long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
long long pow(long long n, long long m, long long mod = 0) {
  if (m < 0) return 0;
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
const int mod = 1000000007;
const int maxn = 55;
long long dp[maxn][maxn][2];
long long c[maxn][maxn];
int n, maxm;
inline void add(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d%d", &n, &maxm);
  memset(dp, 0, sizeof(dp));
  dp[1][0][0] = 1;
  for (int i = 0; i <= n; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      if (c[i][j] >= mod) c[i][j] -= mod;
    }
  }
  for (int i = 2; i <= n; i++)
    for (int j = 0; j <= i - 1; j++) {
      add(dp[i][j][0], dp[i - 1][j][1] * i % mod);
      add(dp[i][j][1], dp[i - 1][j - 1][0] * i % mod);
      for (int k = 1; k <= (i - 1) / 2; k++) {
        int maxt = k / 2;
        for (int t = 0; t <= maxt; t++) {
          long long add0 = dp[k][t][1] * dp[i - k - 1][j - t][1] % mod;
          long long add1 = 0;
          if (t < j) {
            add(add1, dp[k][t][1] * dp[i - k - 1][j - t - 1][0] % mod);
            add(add1, dp[k][t][0] * dp[i - k - 1][j - t - 1][1] % mod);
            add(add1, dp[k][t][0] * dp[i - k - 1][j - t - 1][0] % mod);
          }
          if (k == i - k - 1) {
            add0 *= c[i - 2][k - 1];
            add0 %= mod;
            add1 *= c[i - 2][k - 1];
            add1 %= mod;
          } else {
            add0 *= c[i - 1][k];
            add0 %= mod;
            add1 *= c[i - 1][k];
            add1 %= mod;
          }
          add(dp[i][j][0], add0 * i % mod);
          add(dp[i][j][1], add1 * i % mod);
        }
      }
    }
  cout << (dp[n][maxm][0] + dp[n][maxm][1]) % mod * pow(n, mod - 2, mod) % mod
       << endl;
  return 0;
}

```