#include <bits/stdc++.h>
using namespace std;
const int N = 5e4 + 5;
vector<int> g[N];
void remove(vector<int> &v, int val) {
  for (int i = 0; i < v.size(); i++)
    if (v[i] == val) {
      swap(v[i], v.back());
      v.pop_back();
      i--;
    }
}
void dfs(int node) {
  for (auto &child : g[node]) remove(g[child], node);
  for (auto &child : g[node]) dfs(child);
}
long long dp[N][505];
void solve_path(int node) {
  dp[node][0] = 1;
  for (auto &child : g[node]) solve_path(child);
  for (int want = 1; want <= 500; want++) {
    dp[node][want] = 0;
    for (auto &child : g[node]) dp[node][want] += dp[child][want - 1];
  }
}
long long solve(int node, int want) {
  long long ret = 0;
  long long sum[505];
  memset(sum, 0, sizeof sum);
  for (auto &child : g[node]) {
    for (int j = 0; j <= want - 2; j++) {
      int rem = want - 2 - j;
      ret = (ret + 1ll * dp[child][j] * sum[rem]);
    }
    ret = (ret + dp[child][want - 1]);
    for (int j = 0; j <= 500; j++) sum[j] += dp[child][j];
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  long long ans = 0;
  memset(dp, -1, sizeof dp);
  solve_path(1);
  for (int i = 1; i <= n; i++) ans = (ans + solve(i, k));
  cout << ans;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 int N = 5e4 + 5;
vector<int> g[N];
void remove(vector<int> &v, int val) {
  for (int i = 0; i < v.size(); i++)
    if (v[i] == val) {
      swap(v[i], v.back());
      v.pop_back();
      i--;
    }
}
void dfs(int node) {
  for (auto &child : g[node]) remove(g[child], node);
  for (auto &child : g[node]) dfs(child);
}
long long dp[N][505];
void solve_path(int node) {
  dp[node][0] = 1;
  for (auto &child : g[node]) solve_path(child);
  for (int want = 1; want <= 500; want++) {
    dp[node][want] = 0;
    for (auto &child : g[node]) dp[node][want] += dp[child][want - 1];
  }
}
long long solve(int node, int want) {
  long long ret = 0;
  long long sum[505];
  memset(sum, 0, sizeof sum);
  for (auto &child : g[node]) {
    for (int j = 0; j <= want - 2; j++) {
      int rem = want - 2 - j;
      ret = (ret + 1ll * dp[child][j] * sum[rem]);
    }
    ret = (ret + dp[child][want - 1]);
    for (int j = 0; j <= 500; j++) sum[j] += dp[child][j];
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  long long ans = 0;
  memset(dp, -1, sizeof dp);
  solve_path(1);
  for (int i = 1; i <= n; i++) ans = (ans + solve(i, k));
  cout << ans;
  return 0;
}

```