#include <bits/stdc++.h>
using namespace std;
const int maxN = 5e4 + 7;
vector<int> Adj[maxN];
int n, k, ans = 0;
int f[maxN][501];
void dfs(int u, int p) {
  f[u][0] = 1;
  for (int v : Adj[u]) {
    if (v != p) {
      dfs(v, u);
      for (int i = 1; i <= k; i++) {
        ans += f[u][i - 1] * f[v][k - i];
      }
      for (int i = 1; i <= k; i++) f[u][i] += f[v][i - 1];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    Adj[u].emplace_back(v);
    Adj[v].emplace_back(u);
  }
  dfs(1, 0);
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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 maxN = 5e4 + 7;
vector<int> Adj[maxN];
int n, k, ans = 0;
int f[maxN][501];
void dfs(int u, int p) {
  f[u][0] = 1;
  for (int v : Adj[u]) {
    if (v != p) {
      dfs(v, u);
      for (int i = 1; i <= k; i++) {
        ans += f[u][i - 1] * f[v][k - i];
      }
      for (int i = 1; i <= k; i++) f[u][i] += f[v][i - 1];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    Adj[u].emplace_back(v);
    Adj[v].emplace_back(u);
  }
  dfs(1, 0);
  cout << ans;
  return 0;
}

```