#include <bits/stdc++.h>
using namespace std;
vector<vector<long long int> > v, dp;
long long int ans, k;
void dfs(long long int u, long long int par) {
  long long int i;
  dp[u][0] = 1;
  for (auto it : v[u]) {
    if (it == par) continue;
    dfs(it, u);
    for (i = 1; i <= k; i++) ans += dp[it][i - 1] * dp[u][k - i];
    for (i = 1; i <= k; i++) dp[u][i] += dp[it][i - 1];
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n, i, x, y;
    cin >> n >> k;
    v.resize(n + 1);
    dp.resize(n + 1, vector<long long int>(k + 1));
    ans = 0;
    for (i = 1; i < n; i++) {
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    dfs(1, 0);
    cout << ans << "\n";
  }
  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;
vector<vector<long long int> > v, dp;
long long int ans, k;
void dfs(long long int u, long long int par) {
  long long int i;
  dp[u][0] = 1;
  for (auto it : v[u]) {
    if (it == par) continue;
    dfs(it, u);
    for (i = 1; i <= k; i++) ans += dp[it][i - 1] * dp[u][k - i];
    for (i = 1; i <= k; i++) dp[u][i] += dp[it][i - 1];
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n, i, x, y;
    cin >> n >> k;
    v.resize(n + 1);
    dp.resize(n + 1, vector<long long int>(k + 1));
    ans = 0;
    for (i = 1; i < n; i++) {
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    dfs(1, 0);
    cout << ans << "\n";
  }
  return 0;
}

```