#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:16777216")
vector<int> edges[50000];
int used[50000] = {0};
int dp[50000][500] = {0};
long long cnt = 0;
int k;
void dfs(int n) {
  used[n] = 1;
  dp[n][0] = 1;
  for (int i = 0; i < edges[n].size(); i++) {
    int to = edges[n][i];
    if (!used[to]) {
      dfs(to);
      int to = edges[n][i];
      for (int j = 1; j <= k; j++) {
        cnt += ((long long)dp[n][k - j] * dp[to][j - 1]);
      }
      for (int j = 1; j <= k; j++) dp[n][j] += dp[to][j - 1];
    }
  }
}
int main() {
  int n;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edges[a - 1].push_back(b - 1);
    edges[b - 1].push_back(a - 1);
  }
  dfs(0);
  cout << cnt;
}

### Prompt

Generate a cpp solution to 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;
#pragma comment(linker, "/STACK:16777216")
vector<int> edges[50000];
int used[50000] = {0};
int dp[50000][500] = {0};
long long cnt = 0;
int k;
void dfs(int n) {
  used[n] = 1;
  dp[n][0] = 1;
  for (int i = 0; i < edges[n].size(); i++) {
    int to = edges[n][i];
    if (!used[to]) {
      dfs(to);
      int to = edges[n][i];
      for (int j = 1; j <= k; j++) {
        cnt += ((long long)dp[n][k - j] * dp[to][j - 1]);
      }
      for (int j = 1; j <= k; j++) dp[n][j] += dp[to][j - 1];
    }
  }
}
int main() {
  int n;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edges[a - 1].push_back(b - 1);
    edges[b - 1].push_back(a - 1);
  }
  dfs(0);
  cout << cnt;
}

```