#include <bits/stdc++.h>
using namespace std;
long long cnt0, cnt1, k, n, ans;
long long parent[50004];
long long dp[50005][505];
vector<vector<long long> > tree(50005);
void DFS(long long v, long long p) {
  parent[v] = p;
  dp[v][0] = 1;
  for (long long i = 0; i < tree[v].size(); i++) {
    long long to = tree[v][i];
    if (to == p) continue;
    DFS(to, v);
    for (long long j = 0; j < k; j++) dp[v][j + 1] += dp[to][j];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  DFS(1ll, 0ll);
  for (long long i = 1; i <= n; i++) {
    long long j = i;
    long long counter = k;
    long long t = 0;
    while (parent[j] != 0 && counter >= 0) {
      counter--;
      t++;
      if (counter == 0) {
        cnt0++;
        break;
      } else {
        cnt1 += (dp[parent[j]][k - t] - dp[j][k - t - 1]);
      }
      j = parent[j];
    }
  }
  ans = cnt1 / 2ll + cnt0;
  cout << ans;
  return 0;
}

### Prompt

Generate 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;
long long cnt0, cnt1, k, n, ans;
long long parent[50004];
long long dp[50005][505];
vector<vector<long long> > tree(50005);
void DFS(long long v, long long p) {
  parent[v] = p;
  dp[v][0] = 1;
  for (long long i = 0; i < tree[v].size(); i++) {
    long long to = tree[v][i];
    if (to == p) continue;
    DFS(to, v);
    for (long long j = 0; j < k; j++) dp[v][j + 1] += dp[to][j];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  DFS(1ll, 0ll);
  for (long long i = 1; i <= n; i++) {
    long long j = i;
    long long counter = k;
    long long t = 0;
    while (parent[j] != 0 && counter >= 0) {
      counter--;
      t++;
      if (counter == 0) {
        cnt0++;
        break;
      } else {
        cnt1 += (dp[parent[j]][k - t] - dp[j][k - t - 1]);
      }
      j = parent[j];
    }
  }
  ans = cnt1 / 2ll + cnt0;
  cout << ans;
  return 0;
}

```