#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  long long *g[50000];
  long long szs[50000];
  for (int i = 0; i < 50000; i++) szs[i] = 0;
  int lst[50000][2], st = 0;
  for (int i = 0; i < n - 1; i++) {
    int f, t;
    cin >> f >> t;
    f--;
    t--;
    lst[st][0] = f;
    lst[st++][1] = t;
    szs[f]++;
    szs[t]++;
  }
  for (int i = 0; i < n; i++) g[i] = new long long[szs[i]];
  for (int i = 0; i < 50000; i++) szs[i] = 0;
  for (int i = 0; i < n - 1; i++) {
    g[lst[i][0]][szs[lst[i][0]]++] = lst[i][1];
    g[lst[i][1]][szs[lst[i][1]]++] = lst[i][0];
  }
  long long dp[50001];
  long long sumCN[50001][2];
  for (int i = 0; i < n; i++) {
    dp[i] = sumCN[i][1] = szs[i];
    sumCN[i][0] = 1;
  }
  for (int i = 2; i <= k; i++)
    for (int j = 0; j < n; j++) {
      long long res = 0;
      for (int q = 0; q < szs[j]; q++) {
        int to = g[j][q];
        res += sumCN[to][(i - 1) % 2] - sumCN[j][i % 2];
      }
      sumCN[j][i % 2] += res;
      dp[j] = res;
    }
  long long res = 0;
  for (int i = 0; i < n; i++) res += dp[i];
  cout << res / 2 << endl;
}

### 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>
#pragma comment(linker, "/stack:200000000")
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  long long *g[50000];
  long long szs[50000];
  for (int i = 0; i < 50000; i++) szs[i] = 0;
  int lst[50000][2], st = 0;
  for (int i = 0; i < n - 1; i++) {
    int f, t;
    cin >> f >> t;
    f--;
    t--;
    lst[st][0] = f;
    lst[st++][1] = t;
    szs[f]++;
    szs[t]++;
  }
  for (int i = 0; i < n; i++) g[i] = new long long[szs[i]];
  for (int i = 0; i < 50000; i++) szs[i] = 0;
  for (int i = 0; i < n - 1; i++) {
    g[lst[i][0]][szs[lst[i][0]]++] = lst[i][1];
    g[lst[i][1]][szs[lst[i][1]]++] = lst[i][0];
  }
  long long dp[50001];
  long long sumCN[50001][2];
  for (int i = 0; i < n; i++) {
    dp[i] = sumCN[i][1] = szs[i];
    sumCN[i][0] = 1;
  }
  for (int i = 2; i <= k; i++)
    for (int j = 0; j < n; j++) {
      long long res = 0;
      for (int q = 0; q < szs[j]; q++) {
        int to = g[j][q];
        res += sumCN[to][(i - 1) % 2] - sumCN[j][i % 2];
      }
      sumCN[j][i % 2] += res;
      dp[j] = res;
    }
  long long res = 0;
  for (int i = 0; i < n; i++) res += dp[i];
  cout << res / 2 << endl;
}

```