Problem: You are given an undirected graph with n vertices and m edges. Also, you are given an integer k.

Find either a clique of size k or a non-empty subset of vertices such that each vertex of this subset has at least k neighbors in the subset. If there are no such cliques and subsets report about it.

A subset of vertices is called a clique of size k if its size is k and there exists an edge between every two vertices from the subset. A vertex is called a neighbor of the other vertex if there exists an edge between them.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The next lines contain descriptions of test cases.

The first line of the description of each test case contains three integers n, m, k (1 ≤ n, m, k ≤ 10^5, k ≤ n).

Each of the next m lines contains two integers u, v (1 ≤ u, v ≤ n, u ≠ v), denoting an edge between vertices u and v.

It is guaranteed that there are no self-loops or multiple edges. It is guaranteed that the sum of n for all test cases and the sum of m for all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case: 

If you found a subset of vertices such that each vertex of this subset has at least k neighbors in the subset in the first line output 1 and the size of the subset. On the second line output the vertices of the subset in any order.

If you found a clique of size k then in the first line output 2 and in the second line output the vertices of the clique in any order.

If there are no required subsets and cliques print -1.

If there exists multiple possible answers you can print any of them.

Example

Input


3
5 9 4
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
10 15 3
1 2
2 3
3 4
4 5
5 1
1 7
2 8
3 9
4 10
5 6
7 10
10 8
8 6
6 9
9 7
4 5 4
1 2
2 3
3 4
4 1
1 3


Output


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

Note

In the first test case: the subset \{1, 2, 3, 4\} is a clique of size 4.

In the second test case: degree of each vertex in the original graph is at least 3. So the set of all vertices is a correct answer.

In the third test case: there are no cliques of size 4 or required subsets, so the answer is -1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<unordered_set<int> > gr(n);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    gr[a].insert(b);
    gr[b].insert(a);
  }
  set<int> rem;
  for (int i = 0; i < n; ++i) rem.insert(i);
  map<int, set<int> > degs;
  for (int i = 0; i < n; ++i) degs[gr[i].size()].insert(i);
  while (true) {
    if (rem.size() < k) {
      cout << -1 << endl;
      return;
    }
    if (degs.begin()->first >= k) {
      cout << 1 << " " << rem.size() << endl;
      for (int v : rem) cout << v + 1 << " ";
      cout << endl;
      return;
    }
    int deg_to_remove = degs.begin()->first;
    int to_remove = *degs.begin()->second.begin();
    if (deg_to_remove == k - 1) {
      bool clique = true;
      vector<int> cand;
      cand.push_back(to_remove);
      for (int u : gr[to_remove]) cand.push_back(u);
      for (int i = 0; clique && i < cand.size(); ++i)
        for (int j = i + 1; j < cand.size(); ++j)
          if (!gr[cand[i]].count(cand[j])) {
            clique = false;
            break;
          }
      if (clique) {
        cout << 2 << endl;
        for (int v : cand) cout << v + 1 << " ";
        cout << endl;
        return;
      }
    }
    rem.erase(to_remove);
    for (int u : gr[to_remove]) {
      int deg_u = gr[u].size();
      degs[deg_u].erase(u);
      if (degs[deg_u].empty()) degs.erase(deg_u);
      degs[deg_u - 1].insert(u);
      gr[u].erase(to_remove);
    }
    gr[to_remove].clear();
    if (degs[deg_to_remove].size() == 1)
      degs.erase(deg_to_remove);
    else
      degs[deg_to_remove].erase(to_remove);
  }
}
int main() {
  ios::sync_with_stdio(0);
  int tc;
  cin >> tc;
  for (int t = 0; t < tc; ++t) solve();
  return 0;
}