#include <bits/stdc++.h>
using namespace std;
vector<int> v[50005];
int b[50005][505];
int dfs(int u, int pu, int par[]) {
  par[u] = pu;
  int i;
  for (i = 0; i < v[u].size(); i++) {
    if (v[u][i] != pu) {
      dfs(v[u][i], u, par);
    }
  }
  return 1;
}
int createArray(int u, int k, int pu) {
  if (k == 0) {
    b[u][k] = 1;
    return 1;
  } else if (v[u].size() == 1 && v[u][0] == pu) {
    b[u][k] = 0;
    return 0;
  } else if (b[u][k] != -1)
    return b[u][k];
  else {
    int i, sum = 0;
    for (i = 0; i < v[u].size(); i++) {
      if (v[u][i] != pu) {
        sum += createArray(v[u][i], k - 1, u);
      }
    }
    return b[u][k] = sum;
  }
}
long long nway(int u, int k, int pu) {
  int i, j;
  long long sum = 0;
  for (i = 0; i < v[u].size(); i++) {
    if (v[u][i] != pu) {
      for (j = 1; j <= k - 1; j++) {
        sum += (long long)b[v[u][i]][j - 1] *
               ((long long)b[u][k - j] - (long long)b[v[u][i]][k - j - 1]);
      }
    }
  }
  return sum / 2;
}
int main() {
  int n, i, j, p, q, k;
  scanf("%d", &n);
  scanf("%d", &k);
  for (i = 0; i < n - 1; i++) {
    scanf("%d", &p);
    scanf("%d", &q);
    v[p].push_back(q);
    v[q].push_back(p);
  }
  int par[n + 5];
  dfs(1, 0, par);
  for (i = 0; i < 50005; i++) {
    for (j = 0; j < 505; j++) {
      b[i][j] = -1;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= k; j++) {
      createArray(i, j, par[i]);
    }
  }
  long long sum = 0;
  for (i = 1; i <= n; i++) {
    sum += nway(i, k, par[i]);
    sum += (long long)b[i][k];
  }
  cout << sum << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

A tree is a connected graph that doesn't contain any cycles.

The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.

You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.

Input

The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.

Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.

Output

Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

5 2
1 2
2 3
3 4
2 5


Output

4


Input

5 3
1 2
2 3
3 4
4 5


Output

2

Note

In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> v[50005];
int b[50005][505];
int dfs(int u, int pu, int par[]) {
  par[u] = pu;
  int i;
  for (i = 0; i < v[u].size(); i++) {
    if (v[u][i] != pu) {
      dfs(v[u][i], u, par);
    }
  }
  return 1;
}
int createArray(int u, int k, int pu) {
  if (k == 0) {
    b[u][k] = 1;
    return 1;
  } else if (v[u].size() == 1 && v[u][0] == pu) {
    b[u][k] = 0;
    return 0;
  } else if (b[u][k] != -1)
    return b[u][k];
  else {
    int i, sum = 0;
    for (i = 0; i < v[u].size(); i++) {
      if (v[u][i] != pu) {
        sum += createArray(v[u][i], k - 1, u);
      }
    }
    return b[u][k] = sum;
  }
}
long long nway(int u, int k, int pu) {
  int i, j;
  long long sum = 0;
  for (i = 0; i < v[u].size(); i++) {
    if (v[u][i] != pu) {
      for (j = 1; j <= k - 1; j++) {
        sum += (long long)b[v[u][i]][j - 1] *
               ((long long)b[u][k - j] - (long long)b[v[u][i]][k - j - 1]);
      }
    }
  }
  return sum / 2;
}
int main() {
  int n, i, j, p, q, k;
  scanf("%d", &n);
  scanf("%d", &k);
  for (i = 0; i < n - 1; i++) {
    scanf("%d", &p);
    scanf("%d", &q);
    v[p].push_back(q);
    v[q].push_back(p);
  }
  int par[n + 5];
  dfs(1, 0, par);
  for (i = 0; i < 50005; i++) {
    for (j = 0; j < 505; j++) {
      b[i][j] = -1;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= k; j++) {
      createArray(i, j, par[i]);
    }
  }
  long long sum = 0;
  for (i = 1; i <= n; i++) {
    sum += nway(i, k, par[i]);
    sum += (long long)b[i][k];
  }
  cout << sum << "\n";
  return 0;
}

```