#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
int n, k, f[50111][511], fat[50111];
vector<int> ke[50111];
long long res;
void dfs(int u, int fu) {
  f[u][0] = 1;
  for (int i = 0, _a = (ke[u].size()); i < _a; i++) {
    int v = ke[u][i];
    if (v == fu) continue;
    f[u][1]++;
    fat[v] = u;
    dfs(v, u);
    for (int x = (2), _b = (k); x <= _b; x++) f[u][x] += f[v][x - 1];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  int u, v;
  for (int i = (2), _b = (n); i <= _b; i++) {
    scanf("%d%d", &u, &v);
    ke[u].push_back(v);
    ke[v].push_back(u);
  }
  res = 0;
  dfs(1, -1);
  for (int u = (1), _b = (n); u <= _b; u++) {
    res += f[u][k];
    long long cur = 0;
    for (int i = 0, _a = (ke[u].size()); i < _a; i++) {
      int v = ke[u][i];
      if (v == fat[u]) continue;
      for (int x = (1), _b = (k - 1); x <= _b; x++) {
        cur += f[v][x - 1] * (long long)(f[u][k - x] - f[v][k - x - 1]);
      }
    }
    res += cur / 2;
  }
  cout << res;
  return 0;
}

### Prompt

Your challenge is to write 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 double PI = acos(-1.0);
int n, k, f[50111][511], fat[50111];
vector<int> ke[50111];
long long res;
void dfs(int u, int fu) {
  f[u][0] = 1;
  for (int i = 0, _a = (ke[u].size()); i < _a; i++) {
    int v = ke[u][i];
    if (v == fu) continue;
    f[u][1]++;
    fat[v] = u;
    dfs(v, u);
    for (int x = (2), _b = (k); x <= _b; x++) f[u][x] += f[v][x - 1];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  int u, v;
  for (int i = (2), _b = (n); i <= _b; i++) {
    scanf("%d%d", &u, &v);
    ke[u].push_back(v);
    ke[v].push_back(u);
  }
  res = 0;
  dfs(1, -1);
  for (int u = (1), _b = (n); u <= _b; u++) {
    res += f[u][k];
    long long cur = 0;
    for (int i = 0, _a = (ke[u].size()); i < _a; i++) {
      int v = ke[u][i];
      if (v == fat[u]) continue;
      for (int x = (1), _b = (k - 1); x <= _b; x++) {
        cur += f[v][x - 1] * (long long)(f[u][k - x] - f[v][k - x - 1]);
      }
    }
    res += cur / 2;
  }
  cout << res;
  return 0;
}

```