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;
class graph {
 public:
  int n, m, k;
  set<pair<int, int>> s;
  vector<unordered_set<int>> vec;
  void init() {
    cin >> n >> m >> k;
    s.clear();
    vec.resize(n);
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      vec[u - 1].insert(v - 1);
      vec[v - 1].insert(u - 1);
    }
    for (int i = 0; i < n; i++) {
      s.insert({vec[i].size(), i});
    }
  }
  bool cheak(int ind) {
    vector<int> tmp(vec[ind].begin(), vec[ind].end());
    for (int i = 0; i < tmp.size(); i++) {
      for (int j = i + 1; j < tmp.size(); j++) {
        if (!vec[tmp[i]].count(tmp[j])) return false;
      }
    }
    return true;
  }
  void del(int ind) {
    for (auto i : vec[ind]) {
      s.erase({vec[i].size(), i});
      vec[i].erase(ind);
      s.insert({vec[i].size(), i});
    }
    vec[ind].clear();
    m--;
  }
  void solve() {
    init();
    while (s.size()) {
      int num = s.begin()->first;
      int vertex = s.begin()->second;
      if (num >= k) {
        cout << 1 << ' ' << s.size() << '\n';
        for (auto i : s) cout << i.second + 1 << ' ';
        cout << '\n';
        return;
      }
      if (num == k - 1 && cheak(vertex)) {
        cout << 2 << '\n' << vertex + 1 << ' ';
        for (auto i : vec[vertex]) cout << i + 1 << ' ';
        cout << '\n';
        return;
      }
      s.erase(s.begin());
      del(vertex);
    }
    cout << -1 << '\n';
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    graph g;
    g.solve();
  }
}