#include <bits/stdc++.h>
using namespace std;
const int Max = 50001;
vector<int> edge[Max];
int dp[Max][501];
int n, k;
int dfs(int ro, int fa, int &ans) {
  int dep = 0;
  dp[ro][0] = 1;
  for (int i = 0; i < edge[ro].size(); ++i) {
    if (edge[ro][i] == fa) continue;
    dep = max(dfs(edge[ro][i], ro, ans), dep);
    if (dep > k) dep = k;
    for (int j = 1; j <= k; ++j) ans += dp[ro][j - 1] * dp[edge[ro][i]][k - j];
    for (int j = 1; j <= dep; ++j) dp[ro][j] += dp[edge[ro][i]][j - 1];
  }
  return dep + 1;
}
int main() {
  while (~scanf("%d %d", &n, &k)) {
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < Max; ++i) edge[i].clear();
    for (int i = 0; i < n - 1; ++i) {
      int u, v;
      scanf("%d %d", &u, &v);
      edge[u].push_back(v);
      edge[v].push_back(u);
    }
    int ans = 0;
    dfs(1, -1, ans);
    printf("%d\n", ans);
  }
}

### 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;
const int Max = 50001;
vector<int> edge[Max];
int dp[Max][501];
int n, k;
int dfs(int ro, int fa, int &ans) {
  int dep = 0;
  dp[ro][0] = 1;
  for (int i = 0; i < edge[ro].size(); ++i) {
    if (edge[ro][i] == fa) continue;
    dep = max(dfs(edge[ro][i], ro, ans), dep);
    if (dep > k) dep = k;
    for (int j = 1; j <= k; ++j) ans += dp[ro][j - 1] * dp[edge[ro][i]][k - j];
    for (int j = 1; j <= dep; ++j) dp[ro][j] += dp[edge[ro][i]][j - 1];
  }
  return dep + 1;
}
int main() {
  while (~scanf("%d %d", &n, &k)) {
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < Max; ++i) edge[i].clear();
    for (int i = 0; i < n - 1; ++i) {
      int u, v;
      scanf("%d %d", &u, &v);
      edge[u].push_back(v);
      edge[v].push_back(u);
    }
    int ans = 0;
    dfs(1, -1, ans);
    printf("%d\n", ans);
  }
}

```