#include <bits/stdc++.h>
using namespace std;
const int N = 50500, K = 550;
long long D[N][K];
long long OV[N];
list<int> E[N];
int n, k;
long long ans = 0;
void DFS(int x, int p) {
  D[x][0] = 1;
  for (auto y : E[x]) {
    if (y == p) continue;
    DFS(y, x);
  }
  for (auto y : E[x]) {
    if (y == p) continue;
    for (int d = 1; d <= k; d++) D[x][d] += D[y][d - 1];
  }
  long long add = 0;
  for (int d = 0; d <= k; d++) add += D[x][d] * D[x][k - d];
  for (auto y : E[x]) {
    if (p == y) continue;
    for (int d = 1; d <= k - 1; d++) add -= D[y][d - 1] * D[y][k - d - 1];
  }
  ans += add;
}
int main() {
  cin >> n >> k;
  if (k == 1) {
    cout << n - 1;
    return 0;
  }
  int a, b;
  for (int i = 0; i < n - 1; i++)
    cin >> a >> b, E[a].push_back(b), E[b].push_back(a);
  DFS(1, -1);
  cout << ans / 2;
}

### Prompt

Generate 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;
const int N = 50500, K = 550;
long long D[N][K];
long long OV[N];
list<int> E[N];
int n, k;
long long ans = 0;
void DFS(int x, int p) {
  D[x][0] = 1;
  for (auto y : E[x]) {
    if (y == p) continue;
    DFS(y, x);
  }
  for (auto y : E[x]) {
    if (y == p) continue;
    for (int d = 1; d <= k; d++) D[x][d] += D[y][d - 1];
  }
  long long add = 0;
  for (int d = 0; d <= k; d++) add += D[x][d] * D[x][k - d];
  for (auto y : E[x]) {
    if (p == y) continue;
    for (int d = 1; d <= k - 1; d++) add -= D[y][d - 1] * D[y][k - d - 1];
  }
  ans += add;
}
int main() {
  cin >> n >> k;
  if (k == 1) {
    cout << n - 1;
    return 0;
  }
  int a, b;
  for (int i = 0; i < n - 1; i++)
    cin >> a >> b, E[a].push_back(b), E[b].push_back(a);
  DFS(1, -1);
  cout << ans / 2;
}

```