#include <bits/stdc++.h>
using namespace std;
int N, K;
vector<int> ad[50000];
int an[50000][501];
long long erg = 0;
void visit(int i, int p = -1) {
  for (typeof((ad[i]).begin()) it = (ad[i]).begin(); it != (ad[i]).end();
       it++) {
    if (*it == p) continue;
    visit(*it, i);
    for (int k = 1; k <= K; k++) an[i][k] += an[*it][k - 1];
  }
  an[i][0] = 1;
  erg += 2 * an[i][K];
  long long h = 0;
  for (typeof((ad[i]).begin()) it = (ad[i]).begin(); it != (ad[i]).end();
       it++) {
    if (*it == p) continue;
    for (int k = 1; k < K; k++) {
      h += ((long long)an[*it][k - 1]) * (an[i][K - k] - an[*it][K - k - 1]);
    }
  }
  erg += h;
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    ad[a].push_back(b);
    ad[b].push_back(a);
  }
  visit(0);
  assert(erg % 2 == 0);
  printf("%I64d\n", erg / 2);
  return 0;
}

### Prompt

In CPP, your task is to solve 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, K;
vector<int> ad[50000];
int an[50000][501];
long long erg = 0;
void visit(int i, int p = -1) {
  for (typeof((ad[i]).begin()) it = (ad[i]).begin(); it != (ad[i]).end();
       it++) {
    if (*it == p) continue;
    visit(*it, i);
    for (int k = 1; k <= K; k++) an[i][k] += an[*it][k - 1];
  }
  an[i][0] = 1;
  erg += 2 * an[i][K];
  long long h = 0;
  for (typeof((ad[i]).begin()) it = (ad[i]).begin(); it != (ad[i]).end();
       it++) {
    if (*it == p) continue;
    for (int k = 1; k < K; k++) {
      h += ((long long)an[*it][k - 1]) * (an[i][K - k] - an[*it][K - k - 1]);
    }
  }
  erg += h;
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    ad[a].push_back(b);
    ad[b].push_back(a);
  }
  visit(0);
  assert(erg % 2 == 0);
  printf("%I64d\n", erg / 2);
  return 0;
}

```