#include <bits/stdc++.h>
using namespace std;
vector<int> v[50005];
int A[50005][505];
void dfs(int in, int p) {
  for (int i = 0; i < v[in].size(); i++) {
    if (v[in][i] == p) continue;
    dfs(v[in][i], in);
    for (int j = 1; j < 505; j++) A[in][j] += A[v[in][i]][j - 1];
    A[in][1]++;
  }
}
void dfs1(int in, int p) {
  if (in != 0) {
    int A1[505];
    memset(A1, 0, sizeof(A1));
    for (int i = 1; i < 505; i++) A1[i] = A[p][i] - A[in][i - 1];
    for (int i = 1; i < 505; i++) A[in][i] += A1[i - 1];
    A[in][2]--;
    A[in][1]++;
  }
  for (int i = 0; i < v[in].size(); i++) {
    if (v[in][i] != p) dfs1(v[in][i], in);
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(0, -1);
  dfs1(0, -1);
  long long int sum = 0;
  for (int i = 0; i < n; i++) sum += A[i][k];
  cout << sum / 2 << "\n";
  return 0;
}

### Prompt

Your task is to create 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;
vector<int> v[50005];
int A[50005][505];
void dfs(int in, int p) {
  for (int i = 0; i < v[in].size(); i++) {
    if (v[in][i] == p) continue;
    dfs(v[in][i], in);
    for (int j = 1; j < 505; j++) A[in][j] += A[v[in][i]][j - 1];
    A[in][1]++;
  }
}
void dfs1(int in, int p) {
  if (in != 0) {
    int A1[505];
    memset(A1, 0, sizeof(A1));
    for (int i = 1; i < 505; i++) A1[i] = A[p][i] - A[in][i - 1];
    for (int i = 1; i < 505; i++) A[in][i] += A1[i - 1];
    A[in][2]--;
    A[in][1]++;
  }
  for (int i = 0; i < v[in].size(); i++) {
    if (v[in][i] != p) dfs1(v[in][i], in);
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(0, -1);
  dfs1(0, -1);
  long long int sum = 0;
  for (int i = 0; i < n; i++) sum += A[i][k];
  cout << sum / 2 << "\n";
  return 0;
}

```