#include <bits/stdc++.h>
using namespace std;
int func[50010][520];
void dfs(int u, int p, vector<int> adjlist[], vector<int> &parent,
         vector<bool> visited, int k) {
  parent[u] = p;
  visited[u] = true;
  for (int j = 0; j < adjlist[u].size(); j++) {
    int v = adjlist[u][j];
    if (!visited[v] && v != u) {
      dfs(v, u, adjlist, parent, visited, k);
      for (int i = k; i >= 1; i--) func[u][i] += func[v][i - 1];
    }
  }
  func[u][0] = 1;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  vector<int> adjlist[n + 1], parent(n + 1, -1);
  vector<bool> visited(n + 1, false);
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adjlist[a].push_back(b);
    adjlist[b].push_back(a);
  }
  dfs(1, -1, adjlist, parent, visited, k);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += func[i][k];
    long long val = 0;
    for (auto v : adjlist[i]) {
      if (v != parent[i]) {
        for (int j = 0; j <= k - 2; j++)
          val += func[v][j] * (func[i][k - j - 1] - func[v][k - j - 2]);
      }
    }
    ans += (val / 2);
  }
  cout << ans << "\n";
}

### Prompt

Please create a solution in cpp 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;
int func[50010][520];
void dfs(int u, int p, vector<int> adjlist[], vector<int> &parent,
         vector<bool> visited, int k) {
  parent[u] = p;
  visited[u] = true;
  for (int j = 0; j < adjlist[u].size(); j++) {
    int v = adjlist[u][j];
    if (!visited[v] && v != u) {
      dfs(v, u, adjlist, parent, visited, k);
      for (int i = k; i >= 1; i--) func[u][i] += func[v][i - 1];
    }
  }
  func[u][0] = 1;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  vector<int> adjlist[n + 1], parent(n + 1, -1);
  vector<bool> visited(n + 1, false);
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adjlist[a].push_back(b);
    adjlist[b].push_back(a);
  }
  dfs(1, -1, adjlist, parent, visited, k);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += func[i][k];
    long long val = 0;
    for (auto v : adjlist[i]) {
      if (v != parent[i]) {
        for (int j = 0; j <= k - 2; j++)
          val += func[v][j] * (func[i][k - j - 1] - func[v][k - j - 2]);
      }
    }
    ans += (val / 2);
  }
  cout << ans << "\n";
}

```