#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
template <class T>
typename T::value_type arr_sum(const T& v, int n) {
  typename T::value_type sum = 0;
  for (int i = (0); i < (n); ++i) sum += v[i];
  return sum;
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> g(n - 1);
  for (int i = (0); i < (n - 1); ++i) {
    cin >> g[i].first >> g[i].second;
    --g[i].first;
    --g[i].second;
  }
  vector<vector<long long>> dp(k + 1, vector<long long>(n));
  vector<vector<long long>> dp2(k + 1, vector<long long>(n));
  for (int i = (0); i < (n); ++i) {
    dp2[0][i] = dp[0][i] = 1;
  }
  for (auto i : g) {
    ++dp[1][i.first];
    ++dp[1][i.second];
    ++dp2[1][i.first];
    ++dp2[1][i.second];
  }
  for (int i = (2); i < (k + 1); ++i) {
    for (auto j : g) {
      dp[i][j.second] += dp2[i - 1][j.first];
      dp[i][j.second] -= dp2[i - 2][j.second];
      dp[i][j.first] += dp2[i - 1][j.second];
      dp[i][j.first] -= dp2[i - 2][j.first];
    }
    for (int j = (0); j < (n); ++j) {
      dp2[i][j] = dp[i][j] + dp2[i - 2][j];
    }
  }
  cout << accumulate(dp[k].begin(), dp[k].end(), 0LL) / 2;
}

### 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 MOD = 1000000007;
template <class T>
typename T::value_type arr_sum(const T& v, int n) {
  typename T::value_type sum = 0;
  for (int i = (0); i < (n); ++i) sum += v[i];
  return sum;
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> g(n - 1);
  for (int i = (0); i < (n - 1); ++i) {
    cin >> g[i].first >> g[i].second;
    --g[i].first;
    --g[i].second;
  }
  vector<vector<long long>> dp(k + 1, vector<long long>(n));
  vector<vector<long long>> dp2(k + 1, vector<long long>(n));
  for (int i = (0); i < (n); ++i) {
    dp2[0][i] = dp[0][i] = 1;
  }
  for (auto i : g) {
    ++dp[1][i.first];
    ++dp[1][i.second];
    ++dp2[1][i.first];
    ++dp2[1][i.second];
  }
  for (int i = (2); i < (k + 1); ++i) {
    for (auto j : g) {
      dp[i][j.second] += dp2[i - 1][j.first];
      dp[i][j.second] -= dp2[i - 2][j.second];
      dp[i][j.first] += dp2[i - 1][j.second];
      dp[i][j.first] -= dp2[i - 2][j.first];
    }
    for (int j = (0); j < (n); ++j) {
      dp2[i][j] = dp[i][j] + dp2[i - 2][j];
    }
  }
  cout << accumulate(dp[k].begin(), dp[k].end(), 0LL) / 2;
}

```