#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
const double EPS = 1e-10;
vector<int> G[50010];
vector<int> chil[50010];
int cnt[50010][510];
void dfs(int pos, int par) {
  for (int i = 0; i < (int)G[pos].size(); i++) {
    if (G[pos][i] != par) {
      chil[pos].push_back(G[pos][i]);
      dfs(G[pos][i], pos);
    }
  }
}
void init(int n) {
  memset(cnt, 0, sizeof(cnt));
  dfs(0, -1);
}
void calc(int v, int k, long long &sum) {
  for (int i = 0; i < (int)chil[v].size(); i++) {
    int c = chil[v][i];
    calc(c, k, sum);
    for (int j = 0; j < k; j++) {
      sum += (long long)cnt[c][j] * cnt[v][k - j - 1];
    }
    for (int j = 0; j < k; j++) {
      cnt[v][j + 1] += cnt[c][j];
    }
  }
  cnt[v][0] = 1;
  sum += (long long)cnt[v][0] * cnt[v][k];
}
int main() {
  int n, k, a, b;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    a--, b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  init(n);
  long long res = 0;
  calc(0, k, res);
  cout << res << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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 INF = 1 << 29;
const double EPS = 1e-10;
vector<int> G[50010];
vector<int> chil[50010];
int cnt[50010][510];
void dfs(int pos, int par) {
  for (int i = 0; i < (int)G[pos].size(); i++) {
    if (G[pos][i] != par) {
      chil[pos].push_back(G[pos][i]);
      dfs(G[pos][i], pos);
    }
  }
}
void init(int n) {
  memset(cnt, 0, sizeof(cnt));
  dfs(0, -1);
}
void calc(int v, int k, long long &sum) {
  for (int i = 0; i < (int)chil[v].size(); i++) {
    int c = chil[v][i];
    calc(c, k, sum);
    for (int j = 0; j < k; j++) {
      sum += (long long)cnt[c][j] * cnt[v][k - j - 1];
    }
    for (int j = 0; j < k; j++) {
      cnt[v][j + 1] += cnt[c][j];
    }
  }
  cnt[v][0] = 1;
  sum += (long long)cnt[v][0] * cnt[v][k];
}
int main() {
  int n, k, a, b;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    a--, b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  init(n);
  long long res = 0;
  calc(0, k, res);
  cout << res << endl;
  return 0;
}

```