#include <bits/stdc++.h>
using namespace std;
vector<long long int> adj[50005 + 10];
long long int dp[50005][505];
long long int ans = 0;
void dfs(long long int s, long long int p, long long int k) {
  dp[s][0] = 1;
  for (auto u : adj[s]) {
    if (u != p) {
      dfs(u, s, k);
      for (long long int j = 1; j <= k; j++) {
        ans = ans + dp[s][k - j] * dp[u][j - 1];
      }
      for (long long int j = 1; j <= k; j++) dp[s][j] = dp[s][j] + dp[u][j - 1];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, i, j, k;
  cin >> n >> k;
  for (i = 1; i < n; i++) {
    long long int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, -1, k);
  cout << ans << "\n";
  return (0);
}

### Prompt

In cpp, your task is to solve 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;
vector<long long int> adj[50005 + 10];
long long int dp[50005][505];
long long int ans = 0;
void dfs(long long int s, long long int p, long long int k) {
  dp[s][0] = 1;
  for (auto u : adj[s]) {
    if (u != p) {
      dfs(u, s, k);
      for (long long int j = 1; j <= k; j++) {
        ans = ans + dp[s][k - j] * dp[u][j - 1];
      }
      for (long long int j = 1; j <= k; j++) dp[s][j] = dp[s][j] + dp[u][j - 1];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, i, j, k;
  cin >> n >> k;
  for (i = 1; i < n; i++) {
    long long int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, -1, k);
  cout << ans << "\n";
  return (0);
}

```