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;
int i, j, k, l, n, m, x;
unordered_set<int> a[100009];
set<pair<int, int> > q;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int nt;
  cin >> nt;
  while (nt--) {
    cin >> n >> m >> x;
    for (i = 1; i <= n; i++) a[i].clear();
    for (i = 1; i <= m; i++) {
      cin >> k >> l;
      a[k].insert(l);
      a[l].insert(k);
    }
    q.clear();
    for (i = 1; i <= n; i++) {
      q.insert({(int)a[i].size(), i});
    }
    int flag = 0;
    while (q.size() > 0) {
      pair<int, int> p = *q.begin();
      if (p.first >= x) {
        flag = 1;
        cout << 1 << ' ' << q.size() << '\n';
        for (auto to : q) cout << to.second << ' ';
        cout << '\n';
        break;
      }
      if (p.first == x - 1) {
        vector<int> st;
        for (auto to : a[p.second]) st.push_back(to);
        int can = 1;
        for (auto to : st) {
          for (auto to1 : st) {
            if (to != to1 && a[to].count(to1) == 0) {
              can = 0;
              break;
            }
          }
          if (can == 0) break;
        }
        if (can) {
          cout << 2 << '\n';
          for (auto to : st) cout << to << ' ';
          cout << p.second << '\n';
          flag = 1;
          break;
        }
      }
      if (p.first <= x - 1) {
        for (auto to : a[p.second]) {
          a[to].erase(p.second);
          q.erase({(int)a[to].size() + 1, to});
          if (a[to].size() > 0) q.insert({(int)a[to].size(), to});
        }
        q.erase(p);
      }
    }
    if (!flag) cout << -1 << '\n';
  }
  exit(0);
}