#include <bits/stdc++.h>
using namespace std;
double const EPS = 1e-8, PI = acos(-1);
const int N = 5e4 + 9, M = 3e4 + 9, OO = 1e9 + 1, MOD = 1e9 + 7;
const long long inf = 1e18;
int n, k;
vector<int> vec[N];
long long cnt[N][509], cur;
void DFS(int x, int par) {
  for (int it : vec[x]) {
    if (it == par) continue;
    DFS(it, x);
    for (int i = 0; i < k; ++i) {
      int rem = k - i - 1;
      cur += (long long)cnt[x][i] * cnt[it][rem];
    }
    for (int i = 1; i <= k; ++i) cnt[x][i] += cnt[it][i - 1];
  }
}
int main() {
  cout << fixed << setprecision(0), ios::sync_with_stdio(false),
      cin.tie(nullptr), cout.tie(nullptr);
  cin >> n >> k;
  int u, v;
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  ++n;
  for (int i = 1; i <= n; ++i) cnt[i][0] = 1;
  DFS(1, 0);
  cout << (cur);
  return 0;
}

### 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;
double const EPS = 1e-8, PI = acos(-1);
const int N = 5e4 + 9, M = 3e4 + 9, OO = 1e9 + 1, MOD = 1e9 + 7;
const long long inf = 1e18;
int n, k;
vector<int> vec[N];
long long cnt[N][509], cur;
void DFS(int x, int par) {
  for (int it : vec[x]) {
    if (it == par) continue;
    DFS(it, x);
    for (int i = 0; i < k; ++i) {
      int rem = k - i - 1;
      cur += (long long)cnt[x][i] * cnt[it][rem];
    }
    for (int i = 1; i <= k; ++i) cnt[x][i] += cnt[it][i - 1];
  }
}
int main() {
  cout << fixed << setprecision(0), ios::sync_with_stdio(false),
      cin.tie(nullptr), cout.tie(nullptr);
  cin >> n >> k;
  int u, v;
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  ++n;
  for (int i = 1; i <= n; ++i) cnt[i][0] = 1;
  DFS(1, 0);
  cout << (cur);
  return 0;
}

```