#include <bits/stdc++.h>
using namespace std;
const int M = int(1e5 + 1), INF = 0x3fffffff, mod = 1000000007;
int dp[50009][509], n, k;
long long ans = 0;
vector<int> v[50009];
void dfs(int p, int prev) {
  int u;
  dp[p][0] = 1;
  for (int i = 0; i < v[p].size(); i++) {
    u = v[p][i];
    if (u == prev) continue;
    dfs(u, p);
    for (int i = 0; i < k; i++) ans += dp[p][i] * dp[u][k - i - 1];
    for (int i = 0; i < k; i++) dp[p][i + 1] += dp[u][i];
  }
  return;
}
int main(void) {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, -1);
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int M = int(1e5 + 1), INF = 0x3fffffff, mod = 1000000007;
int dp[50009][509], n, k;
long long ans = 0;
vector<int> v[50009];
void dfs(int p, int prev) {
  int u;
  dp[p][0] = 1;
  for (int i = 0; i < v[p].size(); i++) {
    u = v[p][i];
    if (u == prev) continue;
    dfs(u, p);
    for (int i = 0; i < k; i++) ans += dp[p][i] * dp[u][k - i - 1];
    for (int i = 0; i < k; i++) dp[p][i + 1] += dp[u][i];
  }
  return;
}
int main(void) {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, -1);
  cout << ans << endl;
  return 0;
}

```