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;
typedef priority_queue<long long int> pq_max;
typedef priority_queue<long long int, vector<long long int>,
                       greater<long long int> >
    pq_min;
double pi = 2 * acos(0.0);
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long fast_power(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power % 2 == 1) {
      result = (result * base) % 1000000007;
    }
    base = (base * base) % 1000000007;
    power = power / 2;
  }
  return result;
}
const int N = 200005;
std::vector<int> adj[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int TESTS;
  cin >> TESTS;
  while (TESTS--) {
    int n, k;
    cin >> n >> k;
    int cnt[N], vis[N], sze[N];
    for (long long int i = 0; i < n + 3; i++) {
      adj[i].clear();
      cnt[i] = 0;
      vis[i] = 0;
      sze[i] = 0;
    }
    for (long long int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
      cnt[u]++;
      cnt[v]++;
    }
    std::queue<int> q;
    for (long long int i = 1; i < n + 1; i++) {
      if (cnt[i] == 1) {
        q.push(i);
      }
    }
    int ans = 0;
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      vis[x] = 1;
      int p = adj[x].size();
      for (long long int j = 0; j < p; j++) {
        if (vis[adj[x][j]]) {
          continue;
        }
        cnt[adj[x][j]]--;
        sze[adj[x][j]]++;
        if (sze[adj[x][j]] >= k) {
          ans++;
          sze[adj[x][j]] -= k;
        }
        if (sze[adj[x][j]] % k == 0 && cnt[adj[x][j]] == 1) {
          q.push(adj[x][j]);
        }
      }
    }
    cout << ans << endl;
  }
}
