#include <bits/stdc++.h>
using namespace std;
vector<int> G[50001];
int f[50001][501], k;
long long ans = 0;
void dfs(int vertex, int parent) {
  f[vertex][0] = 1;
  for (auto child : G[vertex]) {
    if (child == parent) continue;
    dfs(child, vertex);
    for (int dist = 1; dist <= k; dist++) {
      f[vertex][dist] += f[child][dist - 1];
    }
  }
  long long soma = 0;
  for (auto child : G[vertex]) {
    if (child == parent) continue;
    for (int i = 1; i < k; i++) {
      if (f[vertex][k - i] == 0) continue;
      soma += f[child][i - 1] * (f[vertex][k - i] - f[child][k - i - 1]);
    }
  }
  ans += soma / 2;
}
int main() {
  ios::sync_with_stdio(false);
  memset(f, 0, sizeof(f));
  int n, u, v;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++) ans += f[i][k];
  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<int> G[50001];
int f[50001][501], k;
long long ans = 0;
void dfs(int vertex, int parent) {
  f[vertex][0] = 1;
  for (auto child : G[vertex]) {
    if (child == parent) continue;
    dfs(child, vertex);
    for (int dist = 1; dist <= k; dist++) {
      f[vertex][dist] += f[child][dist - 1];
    }
  }
  long long soma = 0;
  for (auto child : G[vertex]) {
    if (child == parent) continue;
    for (int i = 1; i < k; i++) {
      if (f[vertex][k - i] == 0) continue;
      soma += f[child][i - 1] * (f[vertex][k - i] - f[child][k - i - 1]);
    }
  }
  ans += soma / 2;
}
int main() {
  ios::sync_with_stdio(false);
  memset(f, 0, sizeof(f));
  int n, u, v;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++) ans += f[i][k];
  cout << ans << '\n';
  return 0;
}

```