#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 10;
const long long mod = 1e9 + 7;
const long long inf = -1e18;
const long long INF = 1e18;
long long us[N], n, k, x, y, dp[50010][510], ans;
vector<long long> g[N];
void dfs(int v) {
  us[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    if (us[g[v][i]] == 0) {
      dfs(g[v][i]);
      for (int j = 1; j < k; j++) {
        if ((dp[g[v][i]][k - j - 1] > 0 || k - j == 1) && dp[v][j] > 0) {
          ans += dp[g[v][i]][k - j - 1] * dp[v][j];
        }
      }
      for (int j = 1; j <= k; j++) {
        dp[v][j] += dp[g[v][i]][j - 1];
      }
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    dp[i][0] = 1;
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    ans += dp[i][k];
  }
  cout << ans;
}

### 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 = 1e5 + 10;
const long long mod = 1e9 + 7;
const long long inf = -1e18;
const long long INF = 1e18;
long long us[N], n, k, x, y, dp[50010][510], ans;
vector<long long> g[N];
void dfs(int v) {
  us[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    if (us[g[v][i]] == 0) {
      dfs(g[v][i]);
      for (int j = 1; j < k; j++) {
        if ((dp[g[v][i]][k - j - 1] > 0 || k - j == 1) && dp[v][j] > 0) {
          ans += dp[g[v][i]][k - j - 1] * dp[v][j];
        }
      }
      for (int j = 1; j <= k; j++) {
        dp[v][j] += dp[g[v][i]][j - 1];
      }
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    dp[i][0] = 1;
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    ans += dp[i][k];
  }
  cout << ans;
}

```