#include <bits/stdc++.h>
using namespace std;
vector<long long> adj[51111];
long long dp[51111][511];
bool b[51111];
long long ans;
long long n, k;
void DFS(long long x) {
  vector<long long> must;
  b[x] = 1;
  for (auto i : adj[x]) {
    if (!b[i]) {
      DFS(i);
      must.push_back(i);
    }
  }
  for (auto i : must) {
    for (long long j = 0; j < k; j++) {
      dp[x][j + 1] += dp[i][j];
    }
  }
  ans += dp[x][k];
  long long tren = 0;
  for (auto i : must) {
    for (long long j = 1; j < k; j++) {
      tren += dp[i][j - 1] * (dp[x][k - j] - dp[i][k - j - 1]);
    }
  }
  ans += tren / 2;
  dp[x][0] = 1;
}
int main() {
  ans = 0;
  scanf("%lld%lld", &n, &k);
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  DFS(1);
  printf("%lld", ans);
  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;
vector<long long> adj[51111];
long long dp[51111][511];
bool b[51111];
long long ans;
long long n, k;
void DFS(long long x) {
  vector<long long> must;
  b[x] = 1;
  for (auto i : adj[x]) {
    if (!b[i]) {
      DFS(i);
      must.push_back(i);
    }
  }
  for (auto i : must) {
    for (long long j = 0; j < k; j++) {
      dp[x][j + 1] += dp[i][j];
    }
  }
  ans += dp[x][k];
  long long tren = 0;
  for (auto i : must) {
    for (long long j = 1; j < k; j++) {
      tren += dp[i][j - 1] * (dp[x][k - j] - dp[i][k - j - 1]);
    }
  }
  ans += tren / 2;
  dp[x][0] = 1;
}
int main() {
  ans = 0;
  scanf("%lld%lld", &n, &k);
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  DFS(1);
  printf("%lld", ans);
  return 0;
}

```