#include <bits/stdc++.h>
using namespace std;
vector<int> vi[50050];
int dp[50050][505];
long long chaeng;
int k;
void dfs(int now, int prev) {
  dp[now][0] = 1;
  for (auto x : vi[now]) {
    if (x == prev) continue;
    dfs(x, now);
    for (int i = 1; i <= k; i++) {
      dp[now][i] += dp[x][i - 1];
    }
  }
  chaeng += dp[now][k];
  long long sum = 0;
  for (auto x : vi[now]) {
    if (x == prev) continue;
    for (int i = 1; i < k; i++) {
      sum += 1LL * dp[x][i - 1] * (dp[now][k - i] - dp[x][k - i - 1]);
    }
  }
  sum /= 2;
  chaeng += sum;
}
int main() {
  int n, i, j, x, y, ans, curr, temp, m;
  cin >> n >> k;
  if (k == 1) {
    cout << n - 1 << endl;
    return 0;
  }
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    vi[x].push_back(y);
    vi[y].push_back(x);
  }
  chaeng = 0;
  dfs(1, 0);
  cout << chaeng << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
vector<int> vi[50050];
int dp[50050][505];
long long chaeng;
int k;
void dfs(int now, int prev) {
  dp[now][0] = 1;
  for (auto x : vi[now]) {
    if (x == prev) continue;
    dfs(x, now);
    for (int i = 1; i <= k; i++) {
      dp[now][i] += dp[x][i - 1];
    }
  }
  chaeng += dp[now][k];
  long long sum = 0;
  for (auto x : vi[now]) {
    if (x == prev) continue;
    for (int i = 1; i < k; i++) {
      sum += 1LL * dp[x][i - 1] * (dp[now][k - i] - dp[x][k - i - 1]);
    }
  }
  sum /= 2;
  chaeng += sum;
}
int main() {
  int n, i, j, x, y, ans, curr, temp, m;
  cin >> n >> k;
  if (k == 1) {
    cout << n - 1 << endl;
    return 0;
  }
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    vi[x].push_back(y);
    vi[y].push_back(x);
  }
  chaeng = 0;
  dfs(1, 0);
  cout << chaeng << endl;
  return 0;
}

```