#include <bits/stdc++.h>
using namespace std;
const int N = 50001;
const int M = 501;
int dp[N][M], lst[N], vf[N * 2], urm[2 * N], nr, ans, k;
void dfs(int x, int pred) {
  dp[x][0] = 1;
  for (int p = lst[x]; p != 0; p = urm[p]) {
    int y = vf[p];
    if (y != pred) {
      dfs(y, x);
      for (int i = 1; i <= k; i++) ans += dp[y][i - 1] * dp[x][k - i];
      for (int i = 1; i <= k; i++) dp[x][i] += dp[y][i - 1];
    }
  }
}
void adauga(int x, int y) {
  vf[++nr] = y;
  urm[nr] = lst[x];
  lst[x] = nr;
}
int main() {
  int n, x, y;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    adauga(x, y);
    adauga(y, x);
  }
  dfs(1, 0);
  cout << ans;
  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;
const int N = 50001;
const int M = 501;
int dp[N][M], lst[N], vf[N * 2], urm[2 * N], nr, ans, k;
void dfs(int x, int pred) {
  dp[x][0] = 1;
  for (int p = lst[x]; p != 0; p = urm[p]) {
    int y = vf[p];
    if (y != pred) {
      dfs(y, x);
      for (int i = 1; i <= k; i++) ans += dp[y][i - 1] * dp[x][k - i];
      for (int i = 1; i <= k; i++) dp[x][i] += dp[y][i - 1];
    }
  }
}
void adauga(int x, int y) {
  vf[++nr] = y;
  urm[nr] = lst[x];
  lst[x] = nr;
}
int main() {
  int n, x, y;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    adauga(x, y);
    adauga(y, x);
  }
  dfs(1, 0);
  cout << ans;
  return 0;
}

```