#include <bits/stdc++.h>
using namespace std;
const int INF = (1LL << 31) - 1;
vector<int> adj[50005];
long long cnt[50005][505];
int n, K;
long long res;
void dfs(int i, int prev) {
  cnt[i][0] = 1;
  for (int j = 0; j < adj[i].size(); ++j)
    if (adj[i][j] != prev) {
      dfs(adj[i][j], i);
      for (int k = 0; k < K; ++k) {
        int k1 = K - k - 1;
        res += cnt[i][k] * cnt[adj[i][j]][k1];
      }
      for (int k = 0; k < K; ++k) cnt[i][k + 1] += cnt[adj[i][j]][k];
    }
}
int main() {
  res = 0;
  scanf("%d %d", &n, &K);
  int u, v;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  memset(cnt, 0, sizeof(cnt));
  dfs(1, -1);
  cout << res << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 INF = (1LL << 31) - 1;
vector<int> adj[50005];
long long cnt[50005][505];
int n, K;
long long res;
void dfs(int i, int prev) {
  cnt[i][0] = 1;
  for (int j = 0; j < adj[i].size(); ++j)
    if (adj[i][j] != prev) {
      dfs(adj[i][j], i);
      for (int k = 0; k < K; ++k) {
        int k1 = K - k - 1;
        res += cnt[i][k] * cnt[adj[i][j]][k1];
      }
      for (int k = 0; k < K; ++k) cnt[i][k + 1] += cnt[adj[i][j]][k];
    }
}
int main() {
  res = 0;
  scanf("%d %d", &n, &K);
  int u, v;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  memset(cnt, 0, sizeof(cnt));
  dfs(1, -1);
  cout << res << endl;
  return 0;
}

```