#include <bits/stdc++.h>
using namespace std;
int n, k, deg[50010];
vector<int> edge[50010];
long long dp[50010][505], cnt;
void call(int u, int par) {
  for (int i = 0; i < ((int)edge[u].size()); i++) {
    int v = edge[u][i];
    if (v == par) continue;
    call(v, u);
    for (int i = 1; i < k; i++) {
      cnt += (dp[v][i - 1] * dp[u][k - i]);
    }
    for (int i = 1; i <= k; i++) {
      dp[u][i] += dp[v][i - 1];
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < (n - 1); i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edge[a - 1].push_back(b - 1);
    edge[b - 1].push_back(a - 1);
    deg[a - 1]++, deg[b - 1]++;
  }
  for (int i = 0; i < (n); i++) {
    dp[i][0] = 1;
    dp[i][0] = 1;
  }
  call(0, -1);
  long long ans = 0;
  for (int i = 0; i < (n); i++) ans += dp[i][k];
  cout << ans + cnt << endl;
  return 0;
}

### 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;
int n, k, deg[50010];
vector<int> edge[50010];
long long dp[50010][505], cnt;
void call(int u, int par) {
  for (int i = 0; i < ((int)edge[u].size()); i++) {
    int v = edge[u][i];
    if (v == par) continue;
    call(v, u);
    for (int i = 1; i < k; i++) {
      cnt += (dp[v][i - 1] * dp[u][k - i]);
    }
    for (int i = 1; i <= k; i++) {
      dp[u][i] += dp[v][i - 1];
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < (n - 1); i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edge[a - 1].push_back(b - 1);
    edge[b - 1].push_back(a - 1);
    deg[a - 1]++, deg[b - 1]++;
  }
  for (int i = 0; i < (n); i++) {
    dp[i][0] = 1;
    dp[i][0] = 1;
  }
  call(0, -1);
  long long ans = 0;
  for (int i = 0; i < (n); i++) ans += dp[i][k];
  cout << ans + cnt << endl;
  return 0;
}

```