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.
#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  set<int> adj[n + 1];
  unordered_set<int> check[n + 1];
  for (long long i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].insert(b);
    adj[b].insert(a);
    check[a].insert(b);
    check[b].insert(a);
  }
  set<pair<int, int> > adjs;
  for (long long i = 1; i < n + 1; i++) adjs.insert({adj[i].size(), i});
  while (!adjs.empty()) {
    auto bro = *(adjs.begin());
    adjs.erase(adjs.begin());
    if (bro.first >= k) {
      vector<int> ans;
      for (long long i = 1; i < n + 1; i++)
        if (adj[i].size() > 0) ans.push_back(i);
      cout << 1 << " " << ans.size() << "\n";
      for (auto j : ans) cout << j << " ";
      cout << "\n";
      return;
    }
    if (bro.first == k - 1) {
      vector<int> ans;
      ans.push_back(bro.second);
      for (auto j : adj[bro.second]) ans.push_back(j);
      assert(ans.size() == k);
      bool hey = true;
      for (long long i = 0; i < k; i++)
        for (long long j = i + 1; j < k; j++)
          if (check[ans[i]].find(ans[j]) == check[ans[i]].end()) {
            hey = false;
            break;
          }
      if (hey) {
        cout << 2 << "\n";
        for (auto j : ans) cout << j << " ";
        cout << "\n";
        return;
      }
    }
    for (auto j : adj[bro.second]) {
      adjs.erase({adj[j].size(), j});
      adj[j].erase(bro.second);
      adjs.insert({adj[j].size(), j});
    }
    adj[bro.second].clear();
  }
  cout << "-1"
       << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
