#include <bits/stdc++.h>
using namespace std;
int mod = int(1e9) + 7;
const int mx = 5 * 1e4 + 7;
vector<int> edges[mx];
const int mxk = 505;
long long int dp[mx][mxk];
int n, k;
long long int ans = 0;
void dfs(int v, int e) {
  dp[v][0]++;
  for (int& h : edges[v]) {
    if (h != e) {
      dfs(h, v);
      for (int i = 0; i <= (k - 1); ++i) {
        ans += (dp[v][i] * dp[h][k - i - 1]);
      }
      for (int i = 1; i <= k; i++) {
        dp[v][i] += dp[h][i - 1];
      }
    }
  }
}
int main() {
  for (int i = 0; i < mx; ++i) {
    for (int j = 0; j < mxk; ++j) {
      dp[i][j] = 0;
    }
  }
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  dfs(1, -1);
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
int mod = int(1e9) + 7;
const int mx = 5 * 1e4 + 7;
vector<int> edges[mx];
const int mxk = 505;
long long int dp[mx][mxk];
int n, k;
long long int ans = 0;
void dfs(int v, int e) {
  dp[v][0]++;
  for (int& h : edges[v]) {
    if (h != e) {
      dfs(h, v);
      for (int i = 0; i <= (k - 1); ++i) {
        ans += (dp[v][i] * dp[h][k - i - 1]);
      }
      for (int i = 1; i <= k; i++) {
        dp[v][i] += dp[h][i - 1];
      }
    }
  }
}
int main() {
  for (int i = 0; i < mx; ++i) {
    for (int j = 0; j < mxk; ++j) {
      dp[i][j] = 0;
    }
  }
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  dfs(1, -1);
  cout << ans << endl;
  return 0;
}

```