#include <bits/stdc++.h>
using namespace std;
int const MAXN = 500 * 100 + 10;
vector<int> adj[MAXN];
int p[MAXN], dp[MAXN][500 + 10], h[MAXN], n, k;
long long ans;
bool vis[MAXN];
void dfs(int i) {
  dp[i][0] = 1;
  vis[i] = true;
  for (int j = 0; j < adj[i].size(); j++) {
    if (!vis[adj[i][j]]) {
      h[adj[i][j]] = h[i] + 1;
      p[adj[i][j]] = i;
      dfs(adj[i][j]);
      for (int l = 1; l < k; l++) {
        dp[i][l] += dp[adj[i][j]][l - 1];
      }
    }
  }
  return;
}
int main() {
  cin >> n >> k;
  int a, b;
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < k; j++) {
      for (int l = 0; l < adj[i].size(); l++) {
        if (p[i] != adj[i][l]) {
          ans += 1ll * dp[adj[i][l]][j - 1] *
                 (dp[i][k - j] - dp[adj[i][l]][k - j - 1]);
        }
      }
    }
  }
  ans /= 2;
  for (int i = 1; i <= n; i++) {
    if (h[i] >= k) ans++;
  }
  cout << ans;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
int const MAXN = 500 * 100 + 10;
vector<int> adj[MAXN];
int p[MAXN], dp[MAXN][500 + 10], h[MAXN], n, k;
long long ans;
bool vis[MAXN];
void dfs(int i) {
  dp[i][0] = 1;
  vis[i] = true;
  for (int j = 0; j < adj[i].size(); j++) {
    if (!vis[adj[i][j]]) {
      h[adj[i][j]] = h[i] + 1;
      p[adj[i][j]] = i;
      dfs(adj[i][j]);
      for (int l = 1; l < k; l++) {
        dp[i][l] += dp[adj[i][j]][l - 1];
      }
    }
  }
  return;
}
int main() {
  cin >> n >> k;
  int a, b;
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < k; j++) {
      for (int l = 0; l < adj[i].size(); l++) {
        if (p[i] != adj[i][l]) {
          ans += 1ll * dp[adj[i][l]][j - 1] *
                 (dp[i][k - j] - dp[adj[i][l]][k - j - 1]);
        }
      }
    }
  }
  ans /= 2;
  for (int i = 1; i <= n; i++) {
    if (h[i] >= k) ans++;
  }
  cout << ans;
  return 0;
}

```