#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > v;
long long ans[50005][505];
long long vis[50005];
long long h[50005];
long long n, k;
long long res = 0;
void dfs(long long i, long long x) {
  vis[i] = 1;
  h[i] = x;
  ans[i][0]++;
  for (auto j : v[i]) {
    if (vis[j] == 0) {
      dfs(j, x + 1);
      for (long long l = 1; l <= k; ++l) {
        ans[i][l] += ans[j][l - 1];
      }
    }
  }
}
void dfs2(long long i) {
  vis[i] = 1;
  res += ans[i][k];
  long long cnt = 0;
  for (auto j : v[i]) {
    if (vis[j] == 0) {
      dfs2(j);
      for (long long l = 0; l <= k; ++l) {
        if (k - l - 2 >= 0)
          cnt += ans[j][l] * (ans[i][k - l - 1] - ans[j][k - l - 2]);
      }
    }
  }
  cnt /= 2;
  res += cnt;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  long long x, y;
  v.resize(n);
  for (long long i = 0; i < n - 1; ++i) {
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(0, 0);
  fill_n(vis, 50005, 0);
  dfs2(0);
  cout << res;
}

### Prompt

Please provide a cpp coded solution 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;
vector<vector<long long> > v;
long long ans[50005][505];
long long vis[50005];
long long h[50005];
long long n, k;
long long res = 0;
void dfs(long long i, long long x) {
  vis[i] = 1;
  h[i] = x;
  ans[i][0]++;
  for (auto j : v[i]) {
    if (vis[j] == 0) {
      dfs(j, x + 1);
      for (long long l = 1; l <= k; ++l) {
        ans[i][l] += ans[j][l - 1];
      }
    }
  }
}
void dfs2(long long i) {
  vis[i] = 1;
  res += ans[i][k];
  long long cnt = 0;
  for (auto j : v[i]) {
    if (vis[j] == 0) {
      dfs2(j);
      for (long long l = 0; l <= k; ++l) {
        if (k - l - 2 >= 0)
          cnt += ans[j][l] * (ans[i][k - l - 1] - ans[j][k - l - 2]);
      }
    }
  }
  cnt /= 2;
  res += cnt;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  long long x, y;
  v.resize(n);
  for (long long i = 0; i < n - 1; ++i) {
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(0, 0);
  fill_n(vis, 50005, 0);
  dfs2(0);
  cout << res;
}

```