#include <bits/stdc++.h>
using namespace std;
const int inf = 2e9;
const double eps = 1e-7;
const double pi = 3.1415926535;
long long d[55555][555];
int n, k;
vector<vector<int> > g;
void dfs1(int v, int p = -1) {
  d[v][0] = 1;
  int size = (int)(g[v]).size();
  for (int i = 0; i < size; i++) {
    int to = g[v][i];
    if (to != p) {
      dfs1(to, v);
      for (int j = 0; j < k; j++) {
        d[v][j + 1] += d[to][j];
      }
    }
  }
}
void dfs2(int v, int p = -1) {
  if (p != -1) {
    for (int j = k - 1; j >= 0; j--) {
      long long rem = (j > 0) ? (d[v][j - 1]) : (0);
      d[v][j + 1] += d[p][j] - rem;
    }
  }
  int size = (int)(g[v]).size();
  for (int i = 0; i < size; i++) {
    int to = g[v][i];
    if (to != p) dfs2(to, v);
  }
}
int main() {
  memset(d, 0, sizeof(d));
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  g.resize(n);
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs1(0);
  dfs2(0);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += d[i][k];
  }
  cout << ans / 2 << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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 inf = 2e9;
const double eps = 1e-7;
const double pi = 3.1415926535;
long long d[55555][555];
int n, k;
vector<vector<int> > g;
void dfs1(int v, int p = -1) {
  d[v][0] = 1;
  int size = (int)(g[v]).size();
  for (int i = 0; i < size; i++) {
    int to = g[v][i];
    if (to != p) {
      dfs1(to, v);
      for (int j = 0; j < k; j++) {
        d[v][j + 1] += d[to][j];
      }
    }
  }
}
void dfs2(int v, int p = -1) {
  if (p != -1) {
    for (int j = k - 1; j >= 0; j--) {
      long long rem = (j > 0) ? (d[v][j - 1]) : (0);
      d[v][j + 1] += d[p][j] - rem;
    }
  }
  int size = (int)(g[v]).size();
  for (int i = 0; i < size; i++) {
    int to = g[v][i];
    if (to != p) dfs2(to, v);
  }
}
int main() {
  memset(d, 0, sizeof(d));
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  g.resize(n);
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs1(0);
  dfs2(0);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += d[i][k];
  }
  cout << ans / 2 << endl;
  return 0;
}

```