#include <bits/stdc++.h>
using namespace std;
long long int d[50001][501];
vector<int> tree[50001];
bool visited[50001];
int k;
void dfs(const int& u, const int& par) {
  d[u][0] = 1;
  visited[u] = true;
  for (auto v : tree[u]) {
    if (!visited[v] && v != par) {
      dfs(v, v);
      for (int h = 1; h <= k; h++) d[u][h] += d[v][h - 1];
    }
  }
}
long long int temp = 0;
long long int dfs2(const int& u, const int& par) {
  visited[u] = true;
  long long int ans = d[u][k];
  temp = 0;
  for (auto v : tree[u])
    if (!visited[v] && v != par) {
      for (int h = 1; h < k; h++)
        temp += (d[v][h - 1] * (d[u][k - h] - d[v][k - h - 1]));
    }
  temp >>= 1;
  ans += temp;
  for (auto v : tree[u])
    if (!visited[v] && v != par) ans += dfs2(v, v);
  return ans;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int n, u, v;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--;
    v--;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  dfs(0, 0);
  memset(visited, false, sizeof(visited));
  cout << dfs2(0, 0) << "\n";
  return 0;
}

### Prompt

Please formulate 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;
long long int d[50001][501];
vector<int> tree[50001];
bool visited[50001];
int k;
void dfs(const int& u, const int& par) {
  d[u][0] = 1;
  visited[u] = true;
  for (auto v : tree[u]) {
    if (!visited[v] && v != par) {
      dfs(v, v);
      for (int h = 1; h <= k; h++) d[u][h] += d[v][h - 1];
    }
  }
}
long long int temp = 0;
long long int dfs2(const int& u, const int& par) {
  visited[u] = true;
  long long int ans = d[u][k];
  temp = 0;
  for (auto v : tree[u])
    if (!visited[v] && v != par) {
      for (int h = 1; h < k; h++)
        temp += (d[v][h - 1] * (d[u][k - h] - d[v][k - h - 1]));
    }
  temp >>= 1;
  ans += temp;
  for (auto v : tree[u])
    if (!visited[v] && v != par) ans += dfs2(v, v);
  return ans;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int n, u, v;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--;
    v--;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  dfs(0, 0);
  memset(visited, false, sizeof(visited));
  cout << dfs2(0, 0) << "\n";
  return 0;
}

```