#include <bits/stdc++.h>
using namespace std;
int n, k;
long long res = 0ll;
vector<vector<long long> > dp;
vector<vector<int> > v;
vector<int> c;
void dfs(int root) {
  c[root] = 0;
  for (int i = 0; i < v[root].size(); ++i)
    if (c[v[root][i]]) {
      dfs(v[root][i]);
      for (int j = 0; j < k; ++j) {
        res = res + dp[root][j] * dp[v[root][i]][k - 1 - j];
      }
      for (int j = 1; j <= k; ++j) dp[root][j] += dp[v[root][i]][j - 1];
    }
}
int main() {
  scanf("%d%d", &n, &k);
  v.resize(n);
  c.resize(n, 1);
  dp.resize(n, vector<long long>(k + 1, 0ll));
  for (int i = 0; i < n; ++i) dp[i][0] = 1;
  for (int i = 0; i + 1 < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    --x, --y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(0);
  cout << res;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
long long res = 0ll;
vector<vector<long long> > dp;
vector<vector<int> > v;
vector<int> c;
void dfs(int root) {
  c[root] = 0;
  for (int i = 0; i < v[root].size(); ++i)
    if (c[v[root][i]]) {
      dfs(v[root][i]);
      for (int j = 0; j < k; ++j) {
        res = res + dp[root][j] * dp[v[root][i]][k - 1 - j];
      }
      for (int j = 1; j <= k; ++j) dp[root][j] += dp[v[root][i]][j - 1];
    }
}
int main() {
  scanf("%d%d", &n, &k);
  v.resize(n);
  c.resize(n, 1);
  dp.resize(n, vector<long long>(k + 1, 0ll));
  for (int i = 0; i < n; ++i) dp[i][0] = 1;
  for (int i = 0; i + 1 < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    --x, --y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(0);
  cout << res;
  return 0;
}

```