#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 50000 + 20;
vector<int> adj[MAX_N];
int dp[MAX_N][500 + 20];
int ans[MAX_N][500 + 20];
int n, k;
long long Sum;
void dfs(int v, int Par = -1) {
  dp[v][0] = 1;
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u != Par) {
      dfs(u, v);
      for (int j = 0; j < k; j++) dp[v][j + 1] += dp[u][j];
    }
  }
  for (int i = 0; i <= k; i++) ans[v][i] = dp[v][i];
}
void fix(int v, int Par = -1) {
  if (Par != -1) {
    for (int i = 0; i < k; i++) {
      ans[v][i + 1] += ans[Par][i];
      if (i - 1 >= 0) ans[v][i + 1] -= dp[v][i - 1];
    }
  }
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u != Par) fix(u, v);
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    --v, u--;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  dfs(0);
  fix(0);
  for (int i = 0; i < n; i++) Sum += ans[i][k];
  cout << Sum / 2 << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int MAX_N = 50000 + 20;
vector<int> adj[MAX_N];
int dp[MAX_N][500 + 20];
int ans[MAX_N][500 + 20];
int n, k;
long long Sum;
void dfs(int v, int Par = -1) {
  dp[v][0] = 1;
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u != Par) {
      dfs(u, v);
      for (int j = 0; j < k; j++) dp[v][j + 1] += dp[u][j];
    }
  }
  for (int i = 0; i <= k; i++) ans[v][i] = dp[v][i];
}
void fix(int v, int Par = -1) {
  if (Par != -1) {
    for (int i = 0; i < k; i++) {
      ans[v][i + 1] += ans[Par][i];
      if (i - 1 >= 0) ans[v][i + 1] -= dp[v][i - 1];
    }
  }
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u != Par) fix(u, v);
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    --v, u--;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  dfs(0);
  fix(0);
  for (int i = 0; i < n; i++) Sum += ans[i][k];
  cout << Sum / 2 << endl;
  return 0;
}

```