#include <bits/stdc++.h>
using namespace std;
int n, k;
vector<int> adj[50005];
vector<int> ls;
int ans = 0, sz[50005];
int cnt[505];
void read() {
  cin >> n >> k;
  for (int i = (1); i <= (n - 1); ++i) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
}
bool is_centroid[50005];
void dfs(int u, int fa) {
  sz[u] = 1;
  for (auto v : adj[u]) {
    if (v == fa || is_centroid[v]) continue;
    dfs(v, u);
    sz[u] += sz[v];
  }
}
int get(int u, int fa, int k) {
  for (auto v : adj[u]) {
    if (v == fa || is_centroid[v]) continue;
    if (sz[v] > k / 2) {
      return get(v, u, k);
    }
  }
  return u;
}
void add(int u, int fa, int kc) {
  if (kc > k) return;
  if (kc == k) ans++;
  ans += cnt[k - kc];
  ls.push_back(kc);
  for (auto v : adj[u]) {
    if (v == fa || is_centroid[v]) continue;
    add(v, u, kc + 1);
  }
}
void sub(int u, int fa, int kc) {
  if (kc > k) return;
  cnt[kc]--;
  for (auto v : adj[u]) {
    if (v == fa || is_centroid[v]) continue;
    sub(v, u, kc + 1);
  }
}
void Cen(int u) {
  dfs(u, 0);
  int centroid = get(u, 0, sz[u]);
  is_centroid[centroid] = true;
  for (auto v : adj[centroid]) {
    if (is_centroid[v]) continue;
    ls.clear();
    add(v, centroid, 1);
    for (auto x : ls) cnt[x]++;
  }
  for (auto v : adj[centroid]) {
    if (is_centroid[v]) continue;
    sub(v, centroid, 1);
  }
  for (auto v : adj[centroid]) {
    if (!is_centroid[v]) Cen(v);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  read();
  Cen(1);
  cout << ans;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int n, k;
vector<int> adj[50005];
vector<int> ls;
int ans = 0, sz[50005];
int cnt[505];
void read() {
  cin >> n >> k;
  for (int i = (1); i <= (n - 1); ++i) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
}
bool is_centroid[50005];
void dfs(int u, int fa) {
  sz[u] = 1;
  for (auto v : adj[u]) {
    if (v == fa || is_centroid[v]) continue;
    dfs(v, u);
    sz[u] += sz[v];
  }
}
int get(int u, int fa, int k) {
  for (auto v : adj[u]) {
    if (v == fa || is_centroid[v]) continue;
    if (sz[v] > k / 2) {
      return get(v, u, k);
    }
  }
  return u;
}
void add(int u, int fa, int kc) {
  if (kc > k) return;
  if (kc == k) ans++;
  ans += cnt[k - kc];
  ls.push_back(kc);
  for (auto v : adj[u]) {
    if (v == fa || is_centroid[v]) continue;
    add(v, u, kc + 1);
  }
}
void sub(int u, int fa, int kc) {
  if (kc > k) return;
  cnt[kc]--;
  for (auto v : adj[u]) {
    if (v == fa || is_centroid[v]) continue;
    sub(v, u, kc + 1);
  }
}
void Cen(int u) {
  dfs(u, 0);
  int centroid = get(u, 0, sz[u]);
  is_centroid[centroid] = true;
  for (auto v : adj[centroid]) {
    if (is_centroid[v]) continue;
    ls.clear();
    add(v, centroid, 1);
    for (auto x : ls) cnt[x]++;
  }
  for (auto v : adj[centroid]) {
    if (is_centroid[v]) continue;
    sub(v, centroid, 1);
  }
  for (auto v : adj[centroid]) {
    if (!is_centroid[v]) Cen(v);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  read();
  Cen(1);
  cout << ans;
  return 0;
}

```