#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:30000000")
using namespace std;
vector<int> g[100000];
vector<int> ch[100000];
bool used[100000];
long long d[50005][505];
void dfs(int v) {
  used[v] = 1;
  bool p = 0;
  for (int(i) = (0); (i) < (g[v].size()); ++(i)) {
    int u = g[v][i];
    if (!used[u]) {
      p = 1;
      dfs(u);
      for (int(j) = (0); (j) < (501); ++(j)) d[v][j + 1] += d[u][j];
      ch[v].push_back(u);
    }
  }
  d[v][0] = 1;
}
long long ans = 0;
int main() {
  int n, k;
  cin >> n >> k;
  for (int(i) = (0); (i) < (n - 1); ++(i)) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  for (int(i) = (0); (i) < (n); ++(i)) {
    ans += d[i][k];
    for (int(j) = (0); (j) < (ch[i].size()); ++(j)) {
      int v = i, u = ch[i][j];
      for (int(z) = (1); (z) < (501); ++(z)) d[v][z] -= d[u][z - 1];
      for (int(z) = (0); (z) < (k); ++(z)) {
        if (min(d[v][z], d[u][k - z - 1]) > 0) {
          ans += d[v][z] * d[u][k - z - 1];
        }
      }
      for (int(z) = (0); (z) < (501); ++(z)) d[v][z + 1] += d[u][z];
    }
  }
  cout << ans / 2ll << endl;
  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>
#pragma comment(linker, "/STACK:30000000")
using namespace std;
vector<int> g[100000];
vector<int> ch[100000];
bool used[100000];
long long d[50005][505];
void dfs(int v) {
  used[v] = 1;
  bool p = 0;
  for (int(i) = (0); (i) < (g[v].size()); ++(i)) {
    int u = g[v][i];
    if (!used[u]) {
      p = 1;
      dfs(u);
      for (int(j) = (0); (j) < (501); ++(j)) d[v][j + 1] += d[u][j];
      ch[v].push_back(u);
    }
  }
  d[v][0] = 1;
}
long long ans = 0;
int main() {
  int n, k;
  cin >> n >> k;
  for (int(i) = (0); (i) < (n - 1); ++(i)) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  for (int(i) = (0); (i) < (n); ++(i)) {
    ans += d[i][k];
    for (int(j) = (0); (j) < (ch[i].size()); ++(j)) {
      int v = i, u = ch[i][j];
      for (int(z) = (1); (z) < (501); ++(z)) d[v][z] -= d[u][z - 1];
      for (int(z) = (0); (z) < (k); ++(z)) {
        if (min(d[v][z], d[u][k - z - 1]) > 0) {
          ans += d[v][z] * d[u][k - z - 1];
        }
      }
      for (int(z) = (0); (z) < (501); ++(z)) d[v][z + 1] += d[u][z];
    }
  }
  cout << ans / 2ll << endl;
  return 0;
}

```