#include <bits/stdc++.h>
using namespace std;
long long inf = (long long)1e18;
long long infint = (int)2e9;
int n, k;
vector<int> G[50005];
vector<int> childs[50005];
long long dp[50005][505];
bool visited[50005];
long long ans = 0;
void dfs(int u) {
  visited[u] = true;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      childs[u].push_back(v);
      dfs(v);
      for (int i = 0; i <= k; i++) dp[u][i + 1] += dp[v][i];
    }
  }
  int ver = u;
  for (int i = 1; i <= (k - i); i++) {
    long long p = 0;
    for (int j = 0; j < childs[ver].size(); j++) {
      p += (dp[childs[ver][j]][i - 1] *
            (dp[ver][k - i] - dp[childs[ver][j]][k - i - 1]));
    }
    if (i == k - i) {
      p /= 2;
    }
    ans += p;
  }
  ans += dp[ver][k];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i <= n; i++) {
    dp[i][0] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    visited[i] = false;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  visited[n - 1] = false, visited[n] = false;
  dfs(1);
  cout << ans << endl;
}

### 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;
long long inf = (long long)1e18;
long long infint = (int)2e9;
int n, k;
vector<int> G[50005];
vector<int> childs[50005];
long long dp[50005][505];
bool visited[50005];
long long ans = 0;
void dfs(int u) {
  visited[u] = true;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      childs[u].push_back(v);
      dfs(v);
      for (int i = 0; i <= k; i++) dp[u][i + 1] += dp[v][i];
    }
  }
  int ver = u;
  for (int i = 1; i <= (k - i); i++) {
    long long p = 0;
    for (int j = 0; j < childs[ver].size(); j++) {
      p += (dp[childs[ver][j]][i - 1] *
            (dp[ver][k - i] - dp[childs[ver][j]][k - i - 1]));
    }
    if (i == k - i) {
      p /= 2;
    }
    ans += p;
  }
  ans += dp[ver][k];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i <= n; i++) {
    dp[i][0] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    visited[i] = false;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  visited[n - 1] = false, visited[n] = false;
  dfs(1);
  cout << ans << endl;
}

```