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;
const int N = 200010;
int T, n, k;
int inq[N], lf[N], sz[N];
vector<int> nex[N];
queue<int> q;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &k);
    while (!q.empty()) q.pop();
    for (int i = 1; i <= n; i++) {
      inq[i] = lf[i] = sz[i] = 0;
      nex[i].clear();
    }
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      nex[u].push_back(v);
      nex[v].push_back(u);
    }
    if (1 == k) {
      printf("%d\n", n - 1);
      continue;
    }
    for (int i = 1; i <= n; i++) {
      sz[i] = nex[i].size();
      if (1 == sz[i]) lf[nex[i][0]] += 1;
    }
    for (int i = 1; i <= n; i++) {
      if (lf[i] < k) continue;
      q.push(i);
      inq[i] = 1;
    }
    int res = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      inq[u] = 0;
      res += 1;
      lf[u] -= k;
      sz[u] -= k;
      if (lf[u] >= k) {
        q.push(u);
        inq[u] = 1;
      }
      if (1 == sz[u]) {
        sz[u] = 0;
        for (int i = 0; i < nex[u].size(); i++) {
          int v = nex[u][i];
          if (0 == sz[v]) continue;
          lf[v] += 1;
          if (lf[v] >= k && !inq[v]) {
            q.push(v);
            inq[v] = 1;
          }
        }
      }
    }
    printf("%d\n", res);
  }
  return 0;
}
