#include <bits/stdc++.h>
using namespace std;
vector<int> G[100001];
int dp[100001][501];
int k;
int sol;
void tree_dp(int x, int px);
int main() {
  int v;
  while (cin >> v >> k) {
    int i, a, b;
    sol = 0;
    memset(dp, false, sizeof(dp));
    for (i = 1; i <= v; i++) G[i].resize(0);
    for (i = 1; i < v; i++) {
      cin >> a >> b;
      G[a].push_back(b);
      G[b].push_back(a);
    }
    tree_dp(1, 1);
    cout << sol << endl;
  }
  return 0;
}
void tree_dp(int x, int px) {
  dp[x][0]++;
  for (int i = 0, len = G[x].size(); i < len; i++) {
    int y = G[x][i];
    if (y == px) continue;
    tree_dp(y, x);
    for (int j = 0; j < k; j++) {
      sol += dp[x][k - j - 1] * dp[y][j];
    }
    for (int j = 1; j <= k; j++) {
      dp[x][j] += dp[y][j - 1];
    }
  }
}

### Prompt

Please formulate a CPP solution 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> G[100001];
int dp[100001][501];
int k;
int sol;
void tree_dp(int x, int px);
int main() {
  int v;
  while (cin >> v >> k) {
    int i, a, b;
    sol = 0;
    memset(dp, false, sizeof(dp));
    for (i = 1; i <= v; i++) G[i].resize(0);
    for (i = 1; i < v; i++) {
      cin >> a >> b;
      G[a].push_back(b);
      G[b].push_back(a);
    }
    tree_dp(1, 1);
    cout << sol << endl;
  }
  return 0;
}
void tree_dp(int x, int px) {
  dp[x][0]++;
  for (int i = 0, len = G[x].size(); i < len; i++) {
    int y = G[x][i];
    if (y == px) continue;
    tree_dp(y, x);
    for (int j = 0; j < k; j++) {
      sol += dp[x][k - j - 1] * dp[y][j];
    }
    for (int j = 1; j <= k; j++) {
      dp[x][j] += dp[y][j - 1];
    }
  }
}

```