#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e4 + 3;
const int maxk = 5e2 + 3;
int n, k;
long long res, f[maxn][maxk], par[maxn];
vector<int> adj[maxn];
inline void Enter() {
  cin >> n >> k;
  for (int u, v, i = 1; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
inline void Dfs(const int &u, const int &p) {
  f[u][0] = 1;
  par[u] = p;
  for (int v : adj[u])
    if (v != p) {
      Dfs(v, u);
      for (int i = 1; i <= k; i++) f[u][i] += f[v][i - 1];
    }
}
inline void Solve() {
  Dfs(1, 0);
  for (int u = 1; u <= n; u++) {
    for (int i = 1; i < k; i++) {
      for (int v : adj[u])
        if (v != par[u]) {
          res += f[v][i - 1] * (f[u][k - i] - f[v][k - i - 1]);
        }
    }
    res += 2 * f[u][k];
  }
  cout << res / 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  if (fopen("TEST"
            ".INP",
            "r"))
    freopen(
        "TEST"
        ".INP",
        "r", stdin),
        freopen(
            "TEST"
            ".OUT",
            "w", stdout);
  auto start = chrono::steady_clock::now();
  Enter();
  Solve();
  auto end = chrono::steady_clock::now();
  cerr << "\nIn milliseconds : "
       << chrono::duration_cast<chrono::milliseconds>(end - start).count();
  cerr << '\n'
       << "In seconds : " << fixed << setprecision(3)
       << (double)chrono::duration_cast<chrono::milliseconds>(end - start)
                  .count() /
              1000
       << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
const int maxn = 5e4 + 3;
const int maxk = 5e2 + 3;
int n, k;
long long res, f[maxn][maxk], par[maxn];
vector<int> adj[maxn];
inline void Enter() {
  cin >> n >> k;
  for (int u, v, i = 1; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
inline void Dfs(const int &u, const int &p) {
  f[u][0] = 1;
  par[u] = p;
  for (int v : adj[u])
    if (v != p) {
      Dfs(v, u);
      for (int i = 1; i <= k; i++) f[u][i] += f[v][i - 1];
    }
}
inline void Solve() {
  Dfs(1, 0);
  for (int u = 1; u <= n; u++) {
    for (int i = 1; i < k; i++) {
      for (int v : adj[u])
        if (v != par[u]) {
          res += f[v][i - 1] * (f[u][k - i] - f[v][k - i - 1]);
        }
    }
    res += 2 * f[u][k];
  }
  cout << res / 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  if (fopen("TEST"
            ".INP",
            "r"))
    freopen(
        "TEST"
        ".INP",
        "r", stdin),
        freopen(
            "TEST"
            ".OUT",
            "w", stdout);
  auto start = chrono::steady_clock::now();
  Enter();
  Solve();
  auto end = chrono::steady_clock::now();
  cerr << "\nIn milliseconds : "
       << chrono::duration_cast<chrono::milliseconds>(end - start).count();
  cerr << '\n'
       << "In seconds : " << fixed << setprecision(3)
       << (double)chrono::duration_cast<chrono::milliseconds>(end - start)
                  .count() /
              1000
       << '\n';
  return 0;
}

```