#include <bits/stdc++.h>
using namespace std;
int n, m, Trace[50000 + 5];
long long d[50000 + 5][500 + 5], res;
vector<int> c[50000 + 5];
void Init();
void Process();
int main() {
  Init();
  Process();
  return 0;
}
void Add(int p, int q) { c[p].push_back(q); }
void Init() {
  int p, q;
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &p, &q);
    Add(p, q);
    Add(q, p);
  }
}
void DFS(int i) {
  vector<int>::iterator it;
  for (it = c[i].begin(); it != c[i].end(); ++it)
    if (Trace[*it] == 0) {
      Trace[*it] = i;
      DFS(*it);
    }
  d[i][0] = 1ll;
  for (it = c[i].begin(); it != c[i].end(); ++it)
    if (Trace[*it] == i) {
      for (int j = 1; j < m; ++j) res += d[i][j] * d[*it][m - 1 - j];
      for (int j = 1; j <= m; ++j) d[i][j] += d[*it][j - 1];
    }
  res += d[i][m];
}
void Process() {
  if (m == 1) {
    printf("%d\n", n - 1);
    return;
  }
  Trace[1] = -1;
  DFS(1);
  cout << res << endl;
}

### 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>
using namespace std;
int n, m, Trace[50000 + 5];
long long d[50000 + 5][500 + 5], res;
vector<int> c[50000 + 5];
void Init();
void Process();
int main() {
  Init();
  Process();
  return 0;
}
void Add(int p, int q) { c[p].push_back(q); }
void Init() {
  int p, q;
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &p, &q);
    Add(p, q);
    Add(q, p);
  }
}
void DFS(int i) {
  vector<int>::iterator it;
  for (it = c[i].begin(); it != c[i].end(); ++it)
    if (Trace[*it] == 0) {
      Trace[*it] = i;
      DFS(*it);
    }
  d[i][0] = 1ll;
  for (it = c[i].begin(); it != c[i].end(); ++it)
    if (Trace[*it] == i) {
      for (int j = 1; j < m; ++j) res += d[i][j] * d[*it][m - 1 - j];
      for (int j = 1; j <= m; ++j) d[i][j] += d[*it][j - 1];
    }
  res += d[i][m];
}
void Process() {
  if (m == 1) {
    printf("%d\n", n - 1);
    return;
  }
  Trace[1] = -1;
  DFS(1);
  cout << res << endl;
}

```