#include <bits/stdc++.h>
using namespace std;
int dp[50005][505];
long long sol = 0;
basic_string<int> e[50005];
int n, k;
void dfs(int x, int p) {
  dp[x][0]++;
  for (int y : e[x]) {
    if (y == p) continue;
    dfs(y, x);
    for (int i = 0; i < k; i++) dp[x][i + 1] += dp[y][i];
    for (int j = 0; j < k - j; j++) {
      if (j == 0 || k - j == 0) continue;
      sol -= dp[y][j - 1] * 1ll * dp[y][k - j - 1];
    }
  }
  for (int j = 0; j < k - j; j++) sol += dp[x][j] * 1ll * dp[x][k - j];
  if (k % 2 == 0) {
    long long extra = dp[x][k >> 1] * 1ll * dp[x][k >> 1];
    for (int y : e[x]) {
      if (y == p) continue;
      extra -= dp[y][(k >> 1) - 1] * 1ll * dp[y][(k >> 1) - 1];
    }
    extra >>= 1;
    sol += extra;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    e[u] += v;
    e[v] += u;
  }
  dfs(1, 1);
  cout << sol << '\n';
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 dp[50005][505];
long long sol = 0;
basic_string<int> e[50005];
int n, k;
void dfs(int x, int p) {
  dp[x][0]++;
  for (int y : e[x]) {
    if (y == p) continue;
    dfs(y, x);
    for (int i = 0; i < k; i++) dp[x][i + 1] += dp[y][i];
    for (int j = 0; j < k - j; j++) {
      if (j == 0 || k - j == 0) continue;
      sol -= dp[y][j - 1] * 1ll * dp[y][k - j - 1];
    }
  }
  for (int j = 0; j < k - j; j++) sol += dp[x][j] * 1ll * dp[x][k - j];
  if (k % 2 == 0) {
    long long extra = dp[x][k >> 1] * 1ll * dp[x][k >> 1];
    for (int y : e[x]) {
      if (y == p) continue;
      extra -= dp[y][(k >> 1) - 1] * 1ll * dp[y][(k >> 1) - 1];
    }
    extra >>= 1;
    sol += extra;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    e[u] += v;
    e[v] += u;
  }
  dfs(1, 1);
  cout << sol << '\n';
}

```