You are given a tree (connected graph without cycles) consisting of n vertices. The tree is unrooted — it is just a connected undirected graph without cycles.

In one move, you can choose exactly k leaves (leaf is such a vertex that is connected to only one another vertex) connected to the same vertex and remove them with edges incident to them. I.e. you choose such leaves u_1, u_2, ..., u_k that there are edges (u_1, v), (u_2, v), ..., (u_k, v) and remove these leaves and these edges.

Your task is to find the maximum number of moves you can perform if you remove leaves optimally.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k < n) — the number of vertices in the tree and the number of leaves you remove in one move, respectively. The next n-1 lines describe edges. The i-th edge is represented as two integers x_i and y_i (1 ≤ x_i, y_i ≤ n), where x_i and y_i are vertices the i-th edge connects. It is guaranteed that the given set of edges forms a tree.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the maximum number of moves you can perform if you remove leaves optimally.

Example

Input


4
8 3
1 2
1 5
7 6
6 8
3 1
6 4
6 1
10 3
1 2
1 10
2 3
1 5
1 6
2 4
7 10
10 9
8 10
7 2
3 1
4 5
3 6
7 4
1 2
1 4
5 1
1 2
2 3
4 3
5 3


Output


2
3
3
4

Note

The picture corresponding to the first test case of the example:

<image>

There you can remove vertices 2, 5 and 3 during the first move and vertices 1, 7 and 4 during the second move.

The picture corresponding to the second test case of the example:

<image>

There you can remove vertices 7, 8 and 9 during the first move, then vertices 5, 6 and 10 during the second move and vertices 1, 3 and 4 during the third move.

The picture corresponding to the third test case of the example:

<image>

There you can remove vertices 5 and 7 during the first move, then vertices 2 and 4 during the second move and vertices 1 and 6 during the third move.
#include <bits/stdc++.h>
using namespace std;
set<int> vi[200200];
int leaf[200200], edge[200200];
void removeEdge(int curr, int take) {
  vector<int> to;
  for (auto x : vi[curr]) {
    if (edge[x] == 1 && take > 0) {
      to.push_back(x);
      take--;
    }
  }
  for (auto x : to) {
    vi[curr].erase(x);
    vi[x].erase(curr);
  }
}
int main() {
  int n, i, j, k, x, y, ans, curr, temp, m;
  int tc;
  cin >> tc;
  while (tc--) {
    scanf("%d %d", &n, &k);
    for (i = 1; i <= n; i++) edge[i] = 0, leaf[i] = 0, vi[i].clear();
    for (i = 1; i < n; i++) {
      scanf("%d %d", &x, &y);
      edge[x]++;
      edge[y]++;
      vi[x].insert(y);
      vi[y].insert(x);
    }
    for (i = 1; i <= n; i++) {
      if (edge[i] == 1) {
        leaf[*vi[i].begin()]++;
      }
    }
    set<pair<int, int>, greater<pair<int, int> > > s;
    for (i = 1; i <= n; i++) {
      s.insert({leaf[i], i});
    }
    ans = 0;
    while (!s.empty()) {
      curr = (*s.begin()).second;
      s.erase(s.begin());
      if (leaf[curr] < k || s.empty()) break;
      int take = leaf[curr] / k;
      ans += take;
      leaf[curr] = leaf[curr] % k;
      edge[curr] -= take * k;
      removeEdge(curr, take * k);
      if (edge[curr] == 1) {
        int to = *vi[curr].begin();
        s.erase({leaf[to], to});
        leaf[to]++;
        s.insert({leaf[to], to});
      } else {
        if (edge[curr] == 0) continue;
        s.insert({leaf[curr], curr});
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
