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 JAVA):

Solution:
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF1440D extends PrintWriter {
	CF1440D() { super(System.out); }
	static class Scanner {
		Scanner(InputStream in) { this.in = in; } InputStream in;
		byte[] bb = new byte[1 << 15]; int i, n;
		byte getc() {
			if (i == n) {
				i = n = 0;
				try { n = in.read(bb); } catch (IOException e) {}
			}
			return i < n ? bb[i++] : 0;
		}
		int nextInt() {
			byte c = 0; while (c <= ' ') c = getc();
			int a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			return a;
		}
	}
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1440D o = new CF1440D(); o.main(); o.flush();
	}

	int[] eo; int[][] ej;
	int[] dd; boolean[] removed;
	int[] qu;
	void init(int n) {
		eo = new int[n]; ej = new int[n][2];
		dd = new int[n]; removed = new boolean[n];
		qu = new int[n];
	}
	void append(int i, int j) {
		int o = eo[i]++; dd[i]++;
		if (o >= 2 && (o & o - 1) == 0)
			ej[i] = Arrays.copyOf(ej[i], o << 1);
		ej[i][o] = j;
	}
	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int m = sc.nextInt();
			int k = sc.nextInt();
			int k_ = (long) k * (k - 1) / 2 <= m ? k - 1 : k;
			init(n);
			HashSet<Long> hs = new HashSet<>();
			for (int h = 0; h < m; h++) {
				int i = sc.nextInt() - 1;
				int j = sc.nextInt() - 1;
				if (i > j) {
					int tmp = i; i = j; j = tmp;
				}
				append(i, j);
				append(j, i);
				hs.add((long) i << 20 | j);
			}
			TreeSet<Integer> pq = new TreeSet<>((i, j) -> dd[i] != dd[j] ? dd[i] - dd[j] : i - j);
			for (int i = 0; i < n; i++)
				pq.add(i);
			boolean found = false;
			while (!pq.isEmpty()) {
				int i = pq.first();
				if (dd[i] >= k) {
					println("1 " + pq.size());
					for (int j : pq)
						print(j + 1 + " ");
					println();
					found = true;
					break;
				}
				if (dd[i] >= k_) {
					boolean clique = true;
					int cnt = 0;
					qu[cnt++] = i;
					for (int o = eo[i]; o-- > 0; ) {
						int j = ej[i][o];
						if (!removed[j])
							qu[cnt++] = j;
					}
out:
					for (int h1 = 1; h1 < cnt; h1++) {
						int u = qu[h1];
						for (int h2 = h1 + 1; h2 < cnt; h2++) {
							int v = qu[h2];
							if (!hs.contains(u < v ? (long) u << 20 | v : (long) v << 20 | u)) {
								clique = false;
								break out;
							}
						}
					}
					if (clique) {
						println(2);
						while (cnt-- > 0)
							print(qu[cnt] + 1 + " ");
						println();
						found = true;
						break;
					}
				}
				pq.remove(i);
				removed[i] = true;
				for (int o = eo[i]; o-- > 0; ) {
					int j = ej[i][o];
					if (!removed[j]) {
						pq.remove(j);
						dd[j]--;
						pq.add(j);
					}
				}
			}
			if (!found)
				println(-1);
		}
	}
}