#include <bits/stdc++.h>
using namespace std;
long long int l, x, y, n, m, i, z, p, s, e, q, j, k, g, b, d, t, h, sum = 0;
string s1, s2;
long long int visited[100009], parent[100009], height[100009];
vector<long long int> v[100009], parentVector[50009];
long long int count1[50009][509];
void dfs(long long int s, long long int h = 1) {
  visited[s] = 1;
  height[s] = h;
  for (auto x : v[s]) {
    if (visited[x] == 0) {
      parent[x] = s;
      dfs(x, h + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y), v[y].push_back(x);
  }
  dfs(1);
  parent[1] = -1;
  for (long long int i = 1; i < n + 1; i++) {
    l = parent[i];
    for (j = 1; j <= 505; j++) {
      if (l == -1) break;
      parentVector[i].push_back(l);
      l = parent[l];
    }
  }
  for (long long int i = 1; i < n + 1; i++) {
    l = 1;
    count1[i][0]++;
    for (auto x : parentVector[i]) {
      count1[x][l]++;
      l++;
    }
  }
  for (long long int i = 1; i < 1; i++) {
    for (long long int j = 0; j < 5; j++) cout << count1[i][j] << " ";
    cout << endl;
  }
  for (long long int i = 1; i < n + 1; i++) {
    l = parent[i];
    long long int prev = i, m = count1[i][k];
    for (j = 1; j < k; j++) {
      if (l == -1) break;
      m += (count1[l][k - j] - count1[prev][k - j - 1]);
      prev = l, l = parent[l];
    }
    if (height[i] > k) {
      m++;
    }
    s += m;
  }
  cout << s / 2;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
long long int l, x, y, n, m, i, z, p, s, e, q, j, k, g, b, d, t, h, sum = 0;
string s1, s2;
long long int visited[100009], parent[100009], height[100009];
vector<long long int> v[100009], parentVector[50009];
long long int count1[50009][509];
void dfs(long long int s, long long int h = 1) {
  visited[s] = 1;
  height[s] = h;
  for (auto x : v[s]) {
    if (visited[x] == 0) {
      parent[x] = s;
      dfs(x, h + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y), v[y].push_back(x);
  }
  dfs(1);
  parent[1] = -1;
  for (long long int i = 1; i < n + 1; i++) {
    l = parent[i];
    for (j = 1; j <= 505; j++) {
      if (l == -1) break;
      parentVector[i].push_back(l);
      l = parent[l];
    }
  }
  for (long long int i = 1; i < n + 1; i++) {
    l = 1;
    count1[i][0]++;
    for (auto x : parentVector[i]) {
      count1[x][l]++;
      l++;
    }
  }
  for (long long int i = 1; i < 1; i++) {
    for (long long int j = 0; j < 5; j++) cout << count1[i][j] << " ";
    cout << endl;
  }
  for (long long int i = 1; i < n + 1; i++) {
    l = parent[i];
    long long int prev = i, m = count1[i][k];
    for (j = 1; j < k; j++) {
      if (l == -1) break;
      m += (count1[l][k - j] - count1[prev][k - j - 1]);
      prev = l, l = parent[l];
    }
    if (height[i] > k) {
      m++;
    }
    s += m;
  }
  cout << s / 2;
}

```