#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<int> gr[50005];
int n, k;
bool vis[50005];
int lev[50005][505];
long long int dp[50005];
void dfs(int v) {
  vis[v] = true;
  lev[v][0] = 1;
  for (auto ed : gr[v]) {
    if (!vis[ed]) {
      dfs(ed);
      for (long long int i = 1; i < k + 1; i++) lev[v][i] += lev[ed][i - 1];
    }
  }
}
void solve(int v) {
  vis[v] = true;
  for (auto ed : gr[v]) {
    if (!vis[ed]) {
      solve(ed);
      for (long long int i = 1; i < k; i++)
        dp[v] += lev[ed][i - 1] * (lev[v][k - i] - lev[ed][k - i - 1]);
    }
  }
  dp[v] /= 2;
  dp[v] += lev[v][k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  int a, b;
  for (long long int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  memset(lev, 0, sizeof(lev));
  memset(vis, false, sizeof(vis));
  dfs(1);
  memset(vis, false, sizeof(vis));
  memset(dp, 0, sizeof(dp));
  solve(1);
  long long int ans = 0;
  for (long long int i = 1; i < n + 1; i++) ans += dp[i];
  cout << ans;
}

### Prompt

Create a solution in Cpp for 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;
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<int> gr[50005];
int n, k;
bool vis[50005];
int lev[50005][505];
long long int dp[50005];
void dfs(int v) {
  vis[v] = true;
  lev[v][0] = 1;
  for (auto ed : gr[v]) {
    if (!vis[ed]) {
      dfs(ed);
      for (long long int i = 1; i < k + 1; i++) lev[v][i] += lev[ed][i - 1];
    }
  }
}
void solve(int v) {
  vis[v] = true;
  for (auto ed : gr[v]) {
    if (!vis[ed]) {
      solve(ed);
      for (long long int i = 1; i < k; i++)
        dp[v] += lev[ed][i - 1] * (lev[v][k - i] - lev[ed][k - i - 1]);
    }
  }
  dp[v] /= 2;
  dp[v] += lev[v][k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  int a, b;
  for (long long int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  memset(lev, 0, sizeof(lev));
  memset(vis, false, sizeof(vis));
  dfs(1);
  memset(vis, false, sizeof(vis));
  memset(dp, 0, sizeof(dp));
  solve(1);
  long long int ans = 0;
  for (long long int i = 1; i < n + 1; i++) ans += dp[i];
  cout << ans;
}

```