#include <bits/stdc++.h>
using namespace std;
vector<int> g[50001];
int dp[50001][510];
int sub[50001];
long long ans = 0;
int k;
void dfs(int v, int dad) {
  dp[v][0] = 1;
  sub[v] = 1;
  for (auto u : g[v]) {
    if (u == dad) continue;
    dfs(u, v);
    sub[v] += sub[u];
    for (int i = 0; i < min(k, sub[v]); i++) {
      ans += dp[v][k - i - 1] * dp[u][i];
    }
    for (int i = 1; i <= min(k, sub[v]); i++) {
      dp[v][i] += dp[u][i - 1];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 dp[50001][510];
int sub[50001];
long long ans = 0;
int k;
void dfs(int v, int dad) {
  dp[v][0] = 1;
  sub[v] = 1;
  for (auto u : g[v]) {
    if (u == dad) continue;
    dfs(u, v);
    sub[v] += sub[u];
    for (int i = 0; i < min(k, sub[v]); i++) {
      ans += dp[v][k - i - 1] * dp[u][i];
    }
    for (int i = 1; i <= min(k, sub[v]); i++) {
      dp[v][i] += dp[u][i - 1];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  cout << ans << '\n';
  return 0;
}

```