#include <bits/stdc++.h>
using namespace std;
long long ans = 0, n, a, b, k, dis[50010][510];
std::vector<long long> v[50010];
void dfs(int u, int p) {
  dis[u][0] = 1;
  for (int i = 0; i < v[u].size(); i++) {
    if (v[u][i] == p) continue;
    dfs(v[u][i], u);
    for (int j = 0; j < k; j++) {
      ans = ans + dis[v[u][i]][j] * dis[u][k - j - 1];
    }
    for (int j = 1; j <= k; j++) {
      dis[u][j] = dis[u][j] + dis[v[u][i]][j - 1];
    }
  }
}
int main() {
  scanf("%lld%lld", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    scanf("%lld%lld", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1, 0);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
long long ans = 0, n, a, b, k, dis[50010][510];
std::vector<long long> v[50010];
void dfs(int u, int p) {
  dis[u][0] = 1;
  for (int i = 0; i < v[u].size(); i++) {
    if (v[u][i] == p) continue;
    dfs(v[u][i], u);
    for (int j = 0; j < k; j++) {
      ans = ans + dis[v[u][i]][j] * dis[u][k - j - 1];
    }
    for (int j = 1; j <= k; j++) {
      dis[u][j] = dis[u][j] + dis[v[u][i]][j - 1];
    }
  }
}
int main() {
  scanf("%lld%lld", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    scanf("%lld%lld", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1, 0);
  printf("%lld\n", ans);
  return 0;
}

```