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;
template <class T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  x *= f;
}
const int N = 2e5 + 5;
int du[N];
struct Node {
  int x, du;
  friend bool operator<(Node a, Node b) {
    return a.du == b.du ? a.x < b.x : a.du > b.du;
  }
} tp;
set<Node> s;
set<Node>::iterator it;
vector<int> e[N];
int n, k, a, b, ct[N], tag[N], ans;
inline void del(int x) { s.erase(s.find((Node){x, ct[x]})); }
inline void ins(int x) { s.insert((Node){x, ct[x]}); }
inline void print() {
  for (Node ys : s) printf("%d %d\n", ys.x, ys.du);
}
inline void topu() {
  for (int i = (1); i <= (n); i++)
    if (du[i] == 1) ct[e[i][0]]++;
  s.clear();
  for (int i = (1); i <= (n); i++)
    if (ct[i] >= k) s.insert((Node){i, ct[i]});
  while (!s.empty()) {
    tp = (*s.begin());
    s.erase(tp);
    int u = tp.x;
    ans += (ct[u] / k);
    du[u] -= (ct[u] - ct[u] % k);
    ct[u] %= k;
    if (du[u] == 1) {
      for (int v : e[u])
        if (du[v] > 1) {
          if (ct[v] >= k) del(v);
          ct[v]++;
          if (ct[v] >= k) ins(v);
        }
    }
  }
  printf("%d\n", ans);
}
inline void solve() {
  read(n);
  read(k);
  ans = 0;
  for (int i = (1); i <= (n); i++) e[i].clear(), du[i] = 0, ct[i] = 0;
  for (int i = (2); i <= (n); i++)
    read(a), read(b), e[a].emplace_back(b), e[b].emplace_back(a), du[a]++,
        du[b]++;
  if (k == 1) return (void)(printf("%d\n", n - 1));
  topu();
}
int main() {
  int TEST;
  read(TEST);
  while (TEST--) solve();
}
