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;
long long M = 1e9 + 7;
double pi = acos(-1.0);
long long powerm(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % M;
    y = y >> 1;
    x = (x * x) % M;
  }
  return res % M;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
const int N = 2e5 + 5;
set<int> adj[N];
int deg[N];
set<int> leaves[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      adj[i].clear();
      leaves[i].clear();
      deg[i] = 0;
    }
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      cin >> u >> v;
      deg[u]++;
      deg[v]++;
      adj[u].insert(v);
      adj[v].insert(u);
    }
    if (k == 1) {
      cout << n - 1 << '\n';
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (deg[i] == 1) {
        leaves[*adj[i].begin()].insert(i);
      }
    }
    set<pair<int, int> > s;
    for (int i = 1; i <= n; i++) {
      int p = -leaves[i].size();
      s.insert(make_pair(p, i));
    }
    int ans = 0;
    while (1) {
      pair<int, int> p = *s.begin();
      int x = -p.first, id = p.second;
      if (x < k) break;
      for (int i = 0; i < k; i++) {
        int g = *leaves[id].begin();
        s.erase(make_pair(-leaves[id].size(), id));
        adj[g].erase(id);
        adj[id].erase(g);
        leaves[id].erase(g);
        deg[g] = 0;
        deg[id]--;
        if (leaves[g].count(id)) leaves[g].erase(id);
        if (adj[id].size() == 1) {
          s.erase(
              make_pair(-leaves[*adj[id].begin()].size(), *adj[id].begin()));
          leaves[*adj[id].begin()].insert(id);
          s.insert(
              make_pair(-leaves[*adj[id].begin()].size(), *adj[id].begin()));
        }
        s.insert(make_pair(-leaves[id].size(), id));
      }
      ans++;
    }
    cout << ans << '\n';
  }
  return 0;
}
