#include <bits/stdc++.h>
using namespace std;
long long n, k, t, res, f[50007][507], tr[50007];
vector<long long> adj[50007];
bool d[50007];
void dfs(long long u) {
  f[u][0] = 1, d[u] = 1;
  long long ans = 0;
  for (auto v : adj[u])
    if (!d[v]) {
      tr[v] = u, dfs(v);
      for (long long i = 1; i <= k; ++i) f[u][i] += f[v][i - 1];
    }
  for (auto v : adj[u]) {
    if (v == tr[u]) continue;
    for (long long i = 1; i < k; ++i)
      ans += (f[v][i - 1]) * (f[u][k - i] - f[v][k - i - 1]);
  }
  res += (ans / 2 + f[u][k]);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i < n; ++i) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v), adj[v].push_back(u);
  }
  dfs(1);
  cout << res;
  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;
long long n, k, t, res, f[50007][507], tr[50007];
vector<long long> adj[50007];
bool d[50007];
void dfs(long long u) {
  f[u][0] = 1, d[u] = 1;
  long long ans = 0;
  for (auto v : adj[u])
    if (!d[v]) {
      tr[v] = u, dfs(v);
      for (long long i = 1; i <= k; ++i) f[u][i] += f[v][i - 1];
    }
  for (auto v : adj[u]) {
    if (v == tr[u]) continue;
    for (long long i = 1; i < k; ++i)
      ans += (f[v][i - 1]) * (f[u][k - i] - f[v][k - i - 1]);
  }
  res += (ans / 2 + f[u][k]);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i < n; ++i) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v), adj[v].push_back(u);
  }
  dfs(1);
  cout << res;
  return 0;
}

```