#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (b == 0) ? a : gcd(b, a % b);
}
vector<vector<long long int> > adj(50005);
vector<long long int> par(50005);
long long int n, k;
void dfs(long long int u, long long int v, vector<vector<long long int> > &d) {
  d[u][0] = 1;
  par[u] = v;
  for (auto i : adj[u]) {
    if (i == v) continue;
    dfs(i, u, d);
    for (long long int j = 1; j <= k; j++) d[u][j] += d[i][j - 1];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (long long int i = 0; i + 1 < n; i++) {
    long long int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vector<vector<long long int> > d(n, vector<long long int>(k + 1, 0));
  dfs(0, -1, d);
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    ans += d[i][k];
    long long int cur = d[i][k];
    cur = 0;
    for (auto j : adj[i]) {
      if (j == par[i]) continue;
      for (long long int x = 1; x < k; x++) {
        cur += (d[j][x - 1] * (d[i][k - x] - d[j][k - x - 1]));
      }
    }
    ans += cur / 2;
  }
  cout << ans << "\n";
}

### 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;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (b == 0) ? a : gcd(b, a % b);
}
vector<vector<long long int> > adj(50005);
vector<long long int> par(50005);
long long int n, k;
void dfs(long long int u, long long int v, vector<vector<long long int> > &d) {
  d[u][0] = 1;
  par[u] = v;
  for (auto i : adj[u]) {
    if (i == v) continue;
    dfs(i, u, d);
    for (long long int j = 1; j <= k; j++) d[u][j] += d[i][j - 1];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (long long int i = 0; i + 1 < n; i++) {
    long long int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vector<vector<long long int> > d(n, vector<long long int>(k + 1, 0));
  dfs(0, -1, d);
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    ans += d[i][k];
    long long int cur = d[i][k];
    cur = 0;
    for (auto j : adj[i]) {
      if (j == par[i]) continue;
      for (long long int x = 1; x < k; x++) {
        cur += (d[j][x - 1] * (d[i][k - x] - d[j][k - x - 1]));
      }
    }
    ans += cur / 2;
  }
  cout << ans << "\n";
}

```