#include <bits/stdc++.h>
using namespace std;
int dp[50000 + 10][500 + 10];
vector<int> grav[50000 + 10];
int parent[50000 + 10];
void dfs(int v, int par) {
  parent[v] = par;
  for (int i = 0; i < (int)grav[v].size(); i++) {
    int to = grav[v][i];
    if (to == par) continue;
    dfs(to, v);
  }
}
int main() {
  ios::sync_with_stdio(false);
  vector<pair<pair<int, int>, int> > evs;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < (int)(n - 1); i++) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    grav[u].push_back(v);
    grav[v].push_back(u);
  }
  dfs(0, -1);
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)grav[i].size(); j++) {
      if (grav[i][j] != parent[i]) {
        ++dp[i][1];
      }
    }
  }
  for (int kk = 1; kk <= k; ++kk) {
    for (int i = 0; i < (int)(n); i++) {
      for (int j = 0; j < (int)grav[i].size(); j++) {
        int to = grav[i][j];
        if (to == parent[i]) continue;
        dp[i][kk] += dp[to][kk - 1];
      }
    }
  }
  for (int i = 0; i < (int)(n); i++) dp[i][0] = 1;
  long long res = 0;
  for (int i = 0; i < (int)(n); i++) {
    int v = i;
    for (int j = 1; j <= k; ++j) {
      int rest = k - j;
      int u = parent[v];
      if (u == -1) break;
      res += dp[u][rest];
      if (rest) {
        res -= dp[v][rest - 1];
      }
      v = u;
    }
  }
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)grav[i].size(); j++) {
      if (grav[i][j] == parent[i]) continue;
      res += dp[grav[i][j]][k - 1];
    }
  }
  cout << res / 2;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int dp[50000 + 10][500 + 10];
vector<int> grav[50000 + 10];
int parent[50000 + 10];
void dfs(int v, int par) {
  parent[v] = par;
  for (int i = 0; i < (int)grav[v].size(); i++) {
    int to = grav[v][i];
    if (to == par) continue;
    dfs(to, v);
  }
}
int main() {
  ios::sync_with_stdio(false);
  vector<pair<pair<int, int>, int> > evs;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < (int)(n - 1); i++) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    grav[u].push_back(v);
    grav[v].push_back(u);
  }
  dfs(0, -1);
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)grav[i].size(); j++) {
      if (grav[i][j] != parent[i]) {
        ++dp[i][1];
      }
    }
  }
  for (int kk = 1; kk <= k; ++kk) {
    for (int i = 0; i < (int)(n); i++) {
      for (int j = 0; j < (int)grav[i].size(); j++) {
        int to = grav[i][j];
        if (to == parent[i]) continue;
        dp[i][kk] += dp[to][kk - 1];
      }
    }
  }
  for (int i = 0; i < (int)(n); i++) dp[i][0] = 1;
  long long res = 0;
  for (int i = 0; i < (int)(n); i++) {
    int v = i;
    for (int j = 1; j <= k; ++j) {
      int rest = k - j;
      int u = parent[v];
      if (u == -1) break;
      res += dp[u][rest];
      if (rest) {
        res -= dp[v][rest - 1];
      }
      v = u;
    }
  }
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)grav[i].size(); j++) {
      if (grav[i][j] == parent[i]) continue;
      res += dp[grav[i][j]][k - 1];
    }
  }
  cout << res / 2;
  return 0;
}

```