#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long n, k;
vector<long long int> v[50005];
long long ans;
vector<long long int> dfs(long long node, long long par) {
  vector<long long int> temp(k + 1);
  temp[0] = 1;
  for (auto i : v[node]) {
    if (i != par) {
      vector<long long int> cc = dfs(i, node);
      for (long long i = 0; i <= k; i++) ans += cc[i] * temp[k - i];
      for (long long i = 0; i <= k; i++) temp[i] += cc[i];
    }
  }
  for (long long i = k; i >= 1; i--) temp[i] = temp[i - 1];
  temp[0] = 0;
  return temp;
}
void solve() {
  cin >> n >> k;
  for (long long i = 1; i <= n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  ans = 0;
  dfs(1, -1);
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    solve();
  }
  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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long n, k;
vector<long long int> v[50005];
long long ans;
vector<long long int> dfs(long long node, long long par) {
  vector<long long int> temp(k + 1);
  temp[0] = 1;
  for (auto i : v[node]) {
    if (i != par) {
      vector<long long int> cc = dfs(i, node);
      for (long long i = 0; i <= k; i++) ans += cc[i] * temp[k - i];
      for (long long i = 0; i <= k; i++) temp[i] += cc[i];
    }
  }
  for (long long i = k; i >= 1; i--) temp[i] = temp[i - 1];
  temp[0] = 0;
  return temp;
}
void solve() {
  cin >> n >> k;
  for (long long i = 1; i <= n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  ans = 0;
  dfs(1, -1);
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    solve();
  }
  return 0;
}

```