#include <bits/stdc++.h>
using namespace std;
const long long N = 5e4 + 1;
const long long K = 500 + 1;
long long n, k;
vector<long long> g[N];
long long dp[N][K];
long long ans = 0;
void dfs(long long v, long long p) {
  dp[v][0] = 1;
  long long tmp = 0;
  for (auto x : g[v]) {
    if (x != p) {
      dfs(x, v);
      for (long long i = 1; i <= k; i++) {
        dp[v][i] += dp[x][i - 1];
      }
    }
  }
  for (auto x : g[v]) {
    if (x != p) {
      for (long long i = 0; i < k - 1; i++) {
        tmp += dp[x][i] * (dp[v][k - i - 1] - dp[x][k - i - 2]);
      }
    }
  }
  dp[v][k] += (tmp >> 1);
}
int main() {
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, -1);
  for (long long i = 1; i <= n; i++) {
    ans += dp[i][k];
  }
  cout << ans << '\n';
}

### 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;
const long long N = 5e4 + 1;
const long long K = 500 + 1;
long long n, k;
vector<long long> g[N];
long long dp[N][K];
long long ans = 0;
void dfs(long long v, long long p) {
  dp[v][0] = 1;
  long long tmp = 0;
  for (auto x : g[v]) {
    if (x != p) {
      dfs(x, v);
      for (long long i = 1; i <= k; i++) {
        dp[v][i] += dp[x][i - 1];
      }
    }
  }
  for (auto x : g[v]) {
    if (x != p) {
      for (long long i = 0; i < k - 1; i++) {
        tmp += dp[x][i] * (dp[v][k - i - 1] - dp[x][k - i - 2]);
      }
    }
  }
  dp[v][k] += (tmp >> 1);
}
int main() {
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, -1);
  for (long long i = 1; i <= n; i++) {
    ans += dp[i][k];
  }
  cout << ans << '\n';
}

```