#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 1;
vector<int> Adj[maxn];
int n, k;
long long res = 0;
bool visited[maxn];
void Input() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    Adj[u].push_back(v);
    Adj[v].push_back(u);
  }
}
vector<int> DFS(int u) {
  vector<int> t(k + 1);
  visited[u] = true;
  t[0] = 1;
  for (int v : Adj[u])
    if (!visited[v]) {
      vector<int> s = DFS(v);
      for (int i = 0; i <= k; i++) res += t[i] * (long long)s[k - i];
      for (int i = 0; i <= k; i++) t[i] += s[i];
    }
  for (int i = k; i >= 1; i--) t[i] = t[i - 1];
  t[0] = 0;
  return t;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fill(visited + 1, visited + 1 + n, false);
  Input();
  DFS(1);
  cout << res;
}

### 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 maxn = 1e5 + 1;
vector<int> Adj[maxn];
int n, k;
long long res = 0;
bool visited[maxn];
void Input() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    Adj[u].push_back(v);
    Adj[v].push_back(u);
  }
}
vector<int> DFS(int u) {
  vector<int> t(k + 1);
  visited[u] = true;
  t[0] = 1;
  for (int v : Adj[u])
    if (!visited[v]) {
      vector<int> s = DFS(v);
      for (int i = 0; i <= k; i++) res += t[i] * (long long)s[k - i];
      for (int i = 0; i <= k; i++) t[i] += s[i];
    }
  for (int i = k; i >= 1; i--) t[i] = t[i - 1];
  t[0] = 0;
  return t;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fill(visited + 1, visited + 1 + n, false);
  Input();
  DFS(1);
  cout << res;
}

```