#include <bits/stdc++.h>
using namespace std;
const int inf = 1000000000;
const int N = 50009;
vector<int> e[N];
int k;
long long f[N][509];
void dfs(int x, int fa) {
  f[x][0] = 1;
  for (int i = 0; i < (int)e[x].size(); i++) {
    int y = e[x][i];
    if (y != fa) {
      dfs(y, x);
      for (int j = 0; j < k; j++) f[x][j + 1] += f[y][j];
    }
  }
}
long long ans = 0;
void dfs2(int x, int fa) {
  ans += 2 * f[x][k];
  for (int i = 0; i < (int)e[x].size(); i++) {
    int y = e[x][i];
    if (y != fa) {
      dfs2(y, x);
      for (int j = 1; j < k; j++)
        ans += f[y][j - 1] * (f[x][k - j] - f[y][k - j - 1]);
    }
  }
}
int main() {
  int n;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(0, -1);
  dfs2(0, -1);
  cout << ans / 2;
}

### Prompt

Your challenge is to write 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;
const int inf = 1000000000;
const int N = 50009;
vector<int> e[N];
int k;
long long f[N][509];
void dfs(int x, int fa) {
  f[x][0] = 1;
  for (int i = 0; i < (int)e[x].size(); i++) {
    int y = e[x][i];
    if (y != fa) {
      dfs(y, x);
      for (int j = 0; j < k; j++) f[x][j + 1] += f[y][j];
    }
  }
}
long long ans = 0;
void dfs2(int x, int fa) {
  ans += 2 * f[x][k];
  for (int i = 0; i < (int)e[x].size(); i++) {
    int y = e[x][i];
    if (y != fa) {
      dfs2(y, x);
      for (int j = 1; j < k; j++)
        ans += f[y][j - 1] * (f[x][k - j] - f[y][k - j - 1]);
    }
  }
}
int main() {
  int n;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(0, -1);
  dfs2(0, -1);
  cout << ans / 2;
}

```