#include <bits/stdc++.h>
using namespace std;
const long long MAX = 1e5 + 99;
multiset<long long> vec[MAX];
long long N, K, X, Y, res, sz, size[MAX], cnt[MAX];
void init(long long node, long long par) {
  size[node] = 1;
  for (auto A : vec[node]) {
    if (A == par) continue;
    init(A, node);
    size[node] += size[A];
  }
  return;
}
long long FindCentroid(long long node, long long par) {
  for (auto A : vec[node]) {
    if (A == par) continue;
    if (size[A] > sz / 2) return FindCentroid(A, node);
  }
  return node;
}
void cal(long long node, long long par, long long hei) {
  if (hei <= K) res += cnt[K - hei];
  for (auto A : vec[node]) {
    if (A == par) continue;
    cal(A, node, hei + 1);
  }
  return;
}
void cal_(long long node, long long par, long long hei, long long delta) {
  cnt[hei] += delta;
  for (auto A : vec[node]) {
    if (A == par) continue;
    cal_(A, node, hei + 1, delta);
  }
  return;
}
void dfs(long long node) {
  long long Node;
  init(node, 0);
  sz = size[node];
  Node = FindCentroid(node, 0);
  cnt[0] = 1;
  for (auto A : vec[Node]) cal(A, Node, 1), cal_(A, Node, 1, 1);
  for (auto A : vec[Node]) cal_(A, Node, 1, -1);
  for (auto A : vec[Node]) vec[A].erase(vec[A].find(Node));
  for (auto A : vec[Node]) dfs(A);
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> K;
  for (long long A = 1; A < N; A++) {
    cin >> X >> Y;
    vec[X].insert(Y);
    vec[Y].insert(X);
  }
  dfs(1);
  cout << res;
  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 long long MAX = 1e5 + 99;
multiset<long long> vec[MAX];
long long N, K, X, Y, res, sz, size[MAX], cnt[MAX];
void init(long long node, long long par) {
  size[node] = 1;
  for (auto A : vec[node]) {
    if (A == par) continue;
    init(A, node);
    size[node] += size[A];
  }
  return;
}
long long FindCentroid(long long node, long long par) {
  for (auto A : vec[node]) {
    if (A == par) continue;
    if (size[A] > sz / 2) return FindCentroid(A, node);
  }
  return node;
}
void cal(long long node, long long par, long long hei) {
  if (hei <= K) res += cnt[K - hei];
  for (auto A : vec[node]) {
    if (A == par) continue;
    cal(A, node, hei + 1);
  }
  return;
}
void cal_(long long node, long long par, long long hei, long long delta) {
  cnt[hei] += delta;
  for (auto A : vec[node]) {
    if (A == par) continue;
    cal_(A, node, hei + 1, delta);
  }
  return;
}
void dfs(long long node) {
  long long Node;
  init(node, 0);
  sz = size[node];
  Node = FindCentroid(node, 0);
  cnt[0] = 1;
  for (auto A : vec[Node]) cal(A, Node, 1), cal_(A, Node, 1, 1);
  for (auto A : vec[Node]) cal_(A, Node, 1, -1);
  for (auto A : vec[Node]) vec[A].erase(vec[A].find(Node));
  for (auto A : vec[Node]) dfs(A);
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> K;
  for (long long A = 1; A < N; A++) {
    cin >> X >> Y;
    vec[X].insert(Y);
    vec[Y].insert(X);
  }
  dfs(1);
  cout << res;
  return 0;
}

```