#include <bits/stdc++.h>
using namespace std;
const int N = 50010;
const int inf = 1e9 + 7;
int n, k;
vector<int> g[N];
long long sum[510], cnt[N][510], ans;
void DP(int u, int fa) {
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (v == fa) continue;
    DP(v, u);
  }
  memset(sum, 0, sizeof sum);
  sum[0] = 1;
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (v == fa) continue;
    for (int j = k - 1; j >= 0; --j) {
      if (cnt[v][j] > 0) cnt[u][j + 1] += cnt[v][j];
      ans += cnt[v][j] * sum[k - j - 1];
    }
    for (int j = 0; j < k; ++j) sum[j + 1] += cnt[v][j];
  }
  cnt[u][0] = 1;
}
void Run() {
  memset(cnt, 0, sizeof cnt);
  for (int i = 0; i <= n; ++i) g[i].clear();
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  ans = 0;
  DP(1, -1);
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  int _, cas = 1;
  while (cin >> n >> k) Run();
}

### 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;
const int N = 50010;
const int inf = 1e9 + 7;
int n, k;
vector<int> g[N];
long long sum[510], cnt[N][510], ans;
void DP(int u, int fa) {
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (v == fa) continue;
    DP(v, u);
  }
  memset(sum, 0, sizeof sum);
  sum[0] = 1;
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (v == fa) continue;
    for (int j = k - 1; j >= 0; --j) {
      if (cnt[v][j] > 0) cnt[u][j + 1] += cnt[v][j];
      ans += cnt[v][j] * sum[k - j - 1];
    }
    for (int j = 0; j < k; ++j) sum[j + 1] += cnt[v][j];
  }
  cnt[u][0] = 1;
}
void Run() {
  memset(cnt, 0, sizeof cnt);
  for (int i = 0; i <= n; ++i) g[i].clear();
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  ans = 0;
  DP(1, -1);
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  int _, cas = 1;
  while (cin >> n >> k) Run();
}

```