#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using i2 = pair<int, int>;
const int N = 5e4 + 9;
vector<int> adj[N];
ll dp1[N][501], dp2[N];
void dfs(int u, int p, int k) {
  int c = 0;
  dp1[u][0] = 1;
  for (int v : adj[u]) {
    if (v != p) {
      dfs(v, u, k);
      c++;
      for (int i = 1; i <= k; i++) dp1[u][i] += dp1[v][i - 1];
    }
  }
  if (c >= 2) {
    for (int v : adj[u]) {
      if (v != p) {
        for (int i = 0; i <= k - 2; i++)
          dp2[u] += dp1[v][i] * (dp1[u][k - i - 1] - dp1[v][k - i - 2]);
      }
    }
    dp2[u] /= 2;
  }
  dp2[u] += dp1[u][k];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  if (k == 1) {
    cout << n - 1;
    return 0;
  }
  dfs(1, 0, k);
  int res = 0;
  for (int i = 1; i <= n; i++) res += dp2[i];
  cout << res;
  return 0;
}

### Prompt

Generate 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;
using ll = long long;
using i2 = pair<int, int>;
const int N = 5e4 + 9;
vector<int> adj[N];
ll dp1[N][501], dp2[N];
void dfs(int u, int p, int k) {
  int c = 0;
  dp1[u][0] = 1;
  for (int v : adj[u]) {
    if (v != p) {
      dfs(v, u, k);
      c++;
      for (int i = 1; i <= k; i++) dp1[u][i] += dp1[v][i - 1];
    }
  }
  if (c >= 2) {
    for (int v : adj[u]) {
      if (v != p) {
        for (int i = 0; i <= k - 2; i++)
          dp2[u] += dp1[v][i] * (dp1[u][k - i - 1] - dp1[v][k - i - 2]);
      }
    }
    dp2[u] /= 2;
  }
  dp2[u] += dp1[u][k];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  if (k == 1) {
    cout << n - 1;
    return 0;
  }
  dfs(1, 0, k);
  int res = 0;
  for (int i = 1; i <= n; i++) res += dp2[i];
  cout << res;
  return 0;
}

```