#include <bits/stdc++.h>
using namespace std;
const int MxN = 5e4;
const int MxK = 5e2;
int dist[MxN + 1][MxK + 1];
vector<int> adj[MxN + 1];
long long kdist[MxN + 1];
void dfs(int child, int parent, int k) {
  for (auto x : adj[child]) {
    if (x != parent) {
      dfs(x, child, k);
      for (int i = 1; i <= k; i++) {
        dist[child][i] += dist[x][i - 1];
      }
    }
  }
  dist[child][0]++;
}
void dfs2(int child, int parent, int k) {
  for (auto x : adj[child]) {
    if (x != parent) {
      dfs2(x, child, k);
      for (int i = 1; i <= k - 1; i++) {
        kdist[child] +=
            (long long)dist[x][i - 1] *
            ((long long)dist[child][k - i] - (long long)dist[x][k - i - 1]);
      }
    }
  }
  kdist[child] /= 2;
  kdist[child] += (long long)dist[child][k];
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0, k);
  dfs2(1, 0, k);
  long long ts = 0;
  for (int i = 1; i <= n; i++) {
    ts += (long long)kdist[i];
  }
  cout << ts << endl;
}

### 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;
const int MxN = 5e4;
const int MxK = 5e2;
int dist[MxN + 1][MxK + 1];
vector<int> adj[MxN + 1];
long long kdist[MxN + 1];
void dfs(int child, int parent, int k) {
  for (auto x : adj[child]) {
    if (x != parent) {
      dfs(x, child, k);
      for (int i = 1; i <= k; i++) {
        dist[child][i] += dist[x][i - 1];
      }
    }
  }
  dist[child][0]++;
}
void dfs2(int child, int parent, int k) {
  for (auto x : adj[child]) {
    if (x != parent) {
      dfs2(x, child, k);
      for (int i = 1; i <= k - 1; i++) {
        kdist[child] +=
            (long long)dist[x][i - 1] *
            ((long long)dist[child][k - i] - (long long)dist[x][k - i - 1]);
      }
    }
  }
  kdist[child] /= 2;
  kdist[child] += (long long)dist[child][k];
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0, k);
  dfs2(1, 0, k);
  long long ts = 0;
  for (int i = 1; i <= n; i++) {
    ts += (long long)kdist[i];
  }
  cout << ts << endl;
}

```