#include <bits/stdc++.h>
using namespace std;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1e9, 1e9);
vector<int> g[100002];
int d[100002], u[100002], k;
long long res, w[505], v[505];
int find(int x, int px, int sz) {
  for (auto& i : g[x])
    if (i != px && !u[i] && d[i] > sz / 2) return find(i, x, sz);
  return x;
}
int dfs(int x, int px) {
  d[x] = 1;
  for (auto& i : g[x])
    if (i != px && !u[i]) d[x] += dfs(i, x);
  return d[x];
}
void dfs1(int x, int px, int dx) {
  if (dx <= k) ++w[dx];
  for (auto& i : g[x])
    if (i != px && !u[i]) dfs1(i, x, dx + 1);
}
void solve(int x) {
  int y = find(x, x, dfs(x, x));
  u[y] = 1;
  fill(v, v + k + 1, 0);
  for (auto& i : g[y]) {
    if (!u[i]) {
      fill(w, w + k + 1, 0);
      dfs1(i, i, 1);
      res += w[k];
      for (int i = 1; i < k; ++i) res += w[i] * v[k - i];
      for (int i = 0; i < k; ++i) v[i] += w[i];
    }
  }
  for (auto& i : g[y])
    if (!u[i]) solve(i);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, x, y;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    cin >> x >> y;
    g[--x].push_back(--y);
    g[y].push_back(x);
  }
  solve(0);
  cout << res << '\n';
}

### 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;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1e9, 1e9);
vector<int> g[100002];
int d[100002], u[100002], k;
long long res, w[505], v[505];
int find(int x, int px, int sz) {
  for (auto& i : g[x])
    if (i != px && !u[i] && d[i] > sz / 2) return find(i, x, sz);
  return x;
}
int dfs(int x, int px) {
  d[x] = 1;
  for (auto& i : g[x])
    if (i != px && !u[i]) d[x] += dfs(i, x);
  return d[x];
}
void dfs1(int x, int px, int dx) {
  if (dx <= k) ++w[dx];
  for (auto& i : g[x])
    if (i != px && !u[i]) dfs1(i, x, dx + 1);
}
void solve(int x) {
  int y = find(x, x, dfs(x, x));
  u[y] = 1;
  fill(v, v + k + 1, 0);
  for (auto& i : g[y]) {
    if (!u[i]) {
      fill(w, w + k + 1, 0);
      dfs1(i, i, 1);
      res += w[k];
      for (int i = 1; i < k; ++i) res += w[i] * v[k - i];
      for (int i = 0; i < k; ++i) v[i] += w[i];
    }
  }
  for (auto& i : g[y])
    if (!u[i]) solve(i);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, x, y;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    cin >> x >> y;
    g[--x].push_back(--y);
    g[y].push_back(x);
  }
  solve(0);
  cout << res << '\n';
}

```