#include <bits/stdc++.h>
using namespace std;
int dX[] = {-1, 0, 1, 0, -1, 1, 1, -1};
int dY[] = {0, 1, 0, -1, 1, 1, -1, -1};
long long int ans;
long long int n, k;
vector<int> adjList[50007];
long long int dp[505][50007];
void dfs(int cur, int par) {
  for (int i = 0; i < adjList[cur].size(); ++i) {
    if (adjList[cur][i] != par) {
      dfs(adjList[cur][i], cur);
    }
  }
  dp[0][cur] = 1;
  for (int j = 1; j <= 500; ++j) {
    for (int i = 0; i < adjList[cur].size(); ++i) {
      if (adjList[cur][i] != par) {
        dp[j][cur] += dp[j - 1][adjList[cur][i]];
      }
    }
  }
  ans += dp[k][cur];
  long long int tmp = 0;
  for (int lft = 1; lft < k; ++lft) {
    for (int i = 0; i < adjList[cur].size(); ++i) {
      if (adjList[cur][i] != par) {
        tmp += (dp[lft - 1][adjList[cur][i]]) *
               (dp[k - lft][cur] - dp[k - lft - 1][adjList[cur][i]]);
      }
    }
  }
  ans += tmp / 2;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long int u, v;
  cin >> n >> k;
  for (long long int i = 1; i < n; ++i) {
    cin >> u >> v;
    adjList[u].push_back(v);
    adjList[v].push_back(u);
  }
  dfs(1, -1);
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int dX[] = {-1, 0, 1, 0, -1, 1, 1, -1};
int dY[] = {0, 1, 0, -1, 1, 1, -1, -1};
long long int ans;
long long int n, k;
vector<int> adjList[50007];
long long int dp[505][50007];
void dfs(int cur, int par) {
  for (int i = 0; i < adjList[cur].size(); ++i) {
    if (adjList[cur][i] != par) {
      dfs(adjList[cur][i], cur);
    }
  }
  dp[0][cur] = 1;
  for (int j = 1; j <= 500; ++j) {
    for (int i = 0; i < adjList[cur].size(); ++i) {
      if (adjList[cur][i] != par) {
        dp[j][cur] += dp[j - 1][adjList[cur][i]];
      }
    }
  }
  ans += dp[k][cur];
  long long int tmp = 0;
  for (int lft = 1; lft < k; ++lft) {
    for (int i = 0; i < adjList[cur].size(); ++i) {
      if (adjList[cur][i] != par) {
        tmp += (dp[lft - 1][adjList[cur][i]]) *
               (dp[k - lft][cur] - dp[k - lft - 1][adjList[cur][i]]);
      }
    }
  }
  ans += tmp / 2;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long int u, v;
  cin >> n >> k;
  for (long long int i = 1; i < n; ++i) {
    cin >> u >> v;
    adjList[u].push_back(v);
    adjList[v].push_back(u);
  }
  dfs(1, -1);
  cout << ans << endl;
  return 0;
}

```