#include <bits/stdc++.h>
using namespace std;
vector<int> adj[100001];
int n;
int k;
long long ans;
vector<int> dfs(int node, int parent = -1) {
  vector<int> fake(k + 1, 0);
  fake[0] = 1;
  for (auto it : adj[node]) {
    if (it == parent) continue;
    vector<int> now = dfs(it, node);
    for (int i = 0; i <= k; i++)
      ans += (long long)fake[i] * (long long)now[k - i];
    for (int i = 0; i <= k; i++) fake[i] += now[i];
  }
  for (int i = k; i >= 1; i--) {
    fake[i] = fake[i - 1];
  }
  fake[0] = 0;
  return fake;
}
void solve() {
  cin >> n;
  cin >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1);
  cout << ans << endl;
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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<int> adj[100001];
int n;
int k;
long long ans;
vector<int> dfs(int node, int parent = -1) {
  vector<int> fake(k + 1, 0);
  fake[0] = 1;
  for (auto it : adj[node]) {
    if (it == parent) continue;
    vector<int> now = dfs(it, node);
    for (int i = 0; i <= k; i++)
      ans += (long long)fake[i] * (long long)now[k - i];
    for (int i = 0; i <= k; i++) fake[i] += now[i];
  }
  for (int i = k; i >= 1; i--) {
    fake[i] = fake[i - 1];
  }
  fake[0] = 0;
  return fake;
}
void solve() {
  cin >> n;
  cin >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1);
  cout << ans << endl;
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}

```