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:
import java.util.*;
import java.io.*;
public class B2 {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		StringBuilder sb = new StringBuilder();
		while(t-- > 0) {
			int n = sc.nextInt();
			int m = sc.nextInt();
			long k = sc.nextInt();
			ArrayList<Integer>[] g = new ArrayList[n];
			HashSet<Integer>[] gcheck = new HashSet[n];
			for(int i = 0; i < n; i++){
				g[i] = new ArrayList<>();
				gcheck[i] = new HashSet<>();
			}
			int[] deg = new int[n];
			for(int i = 0; i < m; i++){
				int u = sc.nextInt()-1;
				int v = sc.nextInt()-1;
				g[u].add(v); g[v].add(u);
				deg[u]++; deg[v]++;
				gcheck[Math.min(u, v)].add(Math.max(u, v));
			}
			if(k * (k-1)/2 > m) {
				sb.append("-1\n"); continue;
			}
			LinkedList<Integer> q = new LinkedList<>();
			boolean[] dead = new boolean[n];
			for(int i = 0; i < n; i++){
				if(deg[i] <= k-1) {
					q.add(i);
				}
			}
			boolean clique = false;
			while(!q.isEmpty()) {
				int u = q.removeFirst();
				if(deg[u] == k-1 && clique(u, g, gcheck, dead)) {
					sb.append(2+"\n");
					sb.append((u+1)+"");
					for(int v: g[u]) {
						if(dead[v]) continue;
						sb.append(" "+(v+1));
					}
					sb.append("\n");
					clique = true;
					break;
				}
				for(int v: g[u]) {
					if(dead[v]) continue;
					deg[v]--;
					if(deg[v] == k-1) {
						q.add(v);
					}
				}
				dead[u] = true;
			}
			if(clique) {
				continue;
			}
			HashSet<Integer> ans = new HashSet<>();
			for(int u = 0; u < n; u++){
				if(!dead[u]) ans.add(u);
			}
			if(ans.size() == 0) {
				sb.append("-1\n"); continue;
			}
			sb.append("1 "+ans.size()+"\n");
			for(int u: ans) {
				sb.append((u+1)+" ");
			}
			sb.replace(sb.length()-1, sb.length(), "\n");
		}
		PrintWriter pw = new PrintWriter(System.out);
		pw.println(sb.toString().trim());
		pw.flush();
		
	}
	static boolean clique(int u, ArrayList<Integer>[] g, HashSet<Integer>[] gcheck, boolean[] dead) {
		for(int i = 0; i < g[u].size(); i++) {
			int v = g[u].get(i);
			if(dead[v]) continue;
			for(int j = i+1; j < g[u].size(); j++) {
				int w = g[u].get(j);
				if(dead[w]) continue;
				if(!gcheck[Math.min(v, w)].contains(Math.max(v, w))) {
					return false;
				}
			}
		}
		return true;
	}
	
	static class FastScanner {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
		public FastScanner() {
			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
			tokenizer = null;
		}
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch(IOException e) {
				throw new RuntimeException(e);
			}
		}
	}

}