#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
const long long MOD = 1e9 + 7;
inline long long add(long long a, long long b, long long m) {
  return (((a % m + b % m) % m + m) % m);
}
inline long long mul(long long a, long long b, long long m) {
  return (((a % m * b % m) % m + m) % m);
}
const int N = 5e4 + 5;
vector<vector<int> > adj(N, vector<int>());
int n, k;
vector<vector<long long> > dp;
void dfs(int u, int p, long long &ans) {
  dp[u][0] = 1;
  vector<long long> res(k + 1, 0);
  for (int v : adj[u]) {
    if (v != p) {
      dfs(v, u, ans);
      for (int i = 1; i < k + 1; i++) {
        dp[u][i] += dp[v][i - 1];
        res[i] += dp[v][i - 1];
      }
    }
  };
  long long temp = 0;
  for (int v : adj[u]) {
    if (v != p) {
      long long temp1 = 0;
      for (int i = 1; i < k + 1; i++) res[i] -= dp[v][i - 1];
      for (int i = 1; i < k; i++) temp1 += dp[v][i - 1] * res[k - i];
      temp += temp1;
    }
  };
  ans += dp[u][k] + temp;
  return;
}
void solve() {
  cin >> n >> k;
  int u, v;
  dp.assign(n + 1, vector<long long>(k + 1, 0));
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long ans = 0;
  dfs(1, 0, ans);
  cout << ans << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int t = 1;
  for (int tc = 1; tc < t + 1; tc++) {
    solve();
  }
  return 0;
}

### Prompt

Generate a CPP solution 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>
#pragma GCC optimize("O2")
using namespace std;
const long long MOD = 1e9 + 7;
inline long long add(long long a, long long b, long long m) {
  return (((a % m + b % m) % m + m) % m);
}
inline long long mul(long long a, long long b, long long m) {
  return (((a % m * b % m) % m + m) % m);
}
const int N = 5e4 + 5;
vector<vector<int> > adj(N, vector<int>());
int n, k;
vector<vector<long long> > dp;
void dfs(int u, int p, long long &ans) {
  dp[u][0] = 1;
  vector<long long> res(k + 1, 0);
  for (int v : adj[u]) {
    if (v != p) {
      dfs(v, u, ans);
      for (int i = 1; i < k + 1; i++) {
        dp[u][i] += dp[v][i - 1];
        res[i] += dp[v][i - 1];
      }
    }
  };
  long long temp = 0;
  for (int v : adj[u]) {
    if (v != p) {
      long long temp1 = 0;
      for (int i = 1; i < k + 1; i++) res[i] -= dp[v][i - 1];
      for (int i = 1; i < k; i++) temp1 += dp[v][i - 1] * res[k - i];
      temp += temp1;
    }
  };
  ans += dp[u][k] + temp;
  return;
}
void solve() {
  cin >> n >> k;
  int u, v;
  dp.assign(n + 1, vector<long long>(k + 1, 0));
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long ans = 0;
  dfs(1, 0, ans);
  cout << ans << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int t = 1;
  for (int tc = 1; tc < t + 1; tc++) {
    solve();
  }
  return 0;
}

```