#include <bits/stdc++.h>
using namespace std;
long long int n, k;
vector<long long int> v[50005];
long long int vis[50005] = {0};
long long int dp[50005][505];
long long int ans = 0;
void dfs(long long int node, long long int parent) {
  vis[node] = 1;
  dp[node][0] = 1;
  long long int si = v[node].size(), i, j;
  for (i = 0; i < si; i++) {
    if (vis[v[node][i]] == 0) {
      dfs(v[node][i], node);
      for (j = 0; j < 504; j++) {
        dp[node][j + 1] += dp[v[node][i]][j];
      }
    }
  }
  ans += dp[node][k];
  long long int temp = 0;
  for (i = 1; i < k; i++) {
    for (j = 0; j < si; j++) {
      if (v[node][j] == parent) continue;
      temp +=
          dp[v[node][j]][i - 1] * (dp[node][k - i] - dp[v[node][j]][k - i - 1]);
    }
  }
  ans += temp / 2;
}
int main() {
  long long int i, j;
  for (i = 0; i < 50005; i++)
    for (j = 0; j < 505; j++) dp[i][j] = 0;
  cin >> n >> k;
  for (i = 0; i < n - 1; i++) {
    long long int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1, 1);
  cout << 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;
long long int n, k;
vector<long long int> v[50005];
long long int vis[50005] = {0};
long long int dp[50005][505];
long long int ans = 0;
void dfs(long long int node, long long int parent) {
  vis[node] = 1;
  dp[node][0] = 1;
  long long int si = v[node].size(), i, j;
  for (i = 0; i < si; i++) {
    if (vis[v[node][i]] == 0) {
      dfs(v[node][i], node);
      for (j = 0; j < 504; j++) {
        dp[node][j + 1] += dp[v[node][i]][j];
      }
    }
  }
  ans += dp[node][k];
  long long int temp = 0;
  for (i = 1; i < k; i++) {
    for (j = 0; j < si; j++) {
      if (v[node][j] == parent) continue;
      temp +=
          dp[v[node][j]][i - 1] * (dp[node][k - i] - dp[v[node][j]][k - i - 1]);
    }
  }
  ans += temp / 2;
}
int main() {
  long long int i, j;
  for (i = 0; i < 50005; i++)
    for (j = 0; j < 505; j++) dp[i][j] = 0;
  cin >> n >> k;
  for (i = 0; i < n - 1; i++) {
    long long int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1, 1);
  cout << ans;
}

```