#include <bits/stdc++.h>
using namespace std;
const int N = 50003, K = 503;
int cnt[K][N];
vector<int> g[N];
long long ans = 0, pans = 0;
int k;
void dfs(int v, int p = -1) {
  cnt[0][v] = 1;
  for (int to : g[v]) {
    if (to != p) {
      dfs(to, v);
      for (int d = 1; d < K; d++) {
        cnt[d][v] += cnt[d - 1][to];
      }
    }
  }
  for (int to : g[v]) {
    if (to == p) continue;
    for (int d = 1; d < k; d++) {
      ans += 1ll * cnt[d - 1][to] * (cnt[k - d][v] - cnt[k - d - 1][to]);
    }
  }
  pans += cnt[k][v];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  cout << ans / 2 + pans;
  return 0;
}

### Prompt

Your task is to create 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 N = 50003, K = 503;
int cnt[K][N];
vector<int> g[N];
long long ans = 0, pans = 0;
int k;
void dfs(int v, int p = -1) {
  cnt[0][v] = 1;
  for (int to : g[v]) {
    if (to != p) {
      dfs(to, v);
      for (int d = 1; d < K; d++) {
        cnt[d][v] += cnt[d - 1][to];
      }
    }
  }
  for (int to : g[v]) {
    if (to == p) continue;
    for (int d = 1; d < k; d++) {
      ans += 1ll * cnt[d - 1][to] * (cnt[k - d][v] - cnt[k - d - 1][to]);
    }
  }
  pans += cnt[k][v];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  cout << ans / 2 + pans;
  return 0;
}

```