1439_B. Graph Subset 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.

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


Example output:
2
4 1 2 3 
1 10
1 2 3 4 5 6 7 8 9 10 
-1


Tags: constructive algorithms
data structures
graphs
Difficulty rating: 2600

Solution:
from __future__ import division,print_function;from heapq import *;import sys
le = sys.__stdin__.read().split("\n")[::-1];af = []
for ezo in range(int(le.pop())):
    n,m,k=list(map(int,le.pop().split()));ar=[set() for i in range(n)]
    for i in range(m):a,b=map(int,le.pop().split());ar[a-1].add(b-1);ar[b-1].add(a-1)
    if k*(k-1)>2*m:        af.append(-1)
    else:
        pi=[i for i in range(n) if len(ar[i])<k];re=set(i for i in range(n) if len(ar[i])>=k)
        while pi:
            a=pi.pop()
            if len(ar[a])==k-1:
                clique=True
                for j in ar[a]:
                    if clique:
                        for h in ar[a]:
                            if not(h in ar[j]) and h!=j:clique=False;break
                if clique:af.append(2);af.append(" ".join(map(str,[i+1 for i in ar[a]]+[a+1])));pi=[];re={-1}
            for j in ar[a]:
                ar[j].remove(a)
                if len(ar[j])<k:
                    if j in re:pi.append(j);re.discard(j)
        if re and re!={-1}:af.append("1"+" "+str(len(re)));af.append(" ".join(map(str,[i+1 for i in re])))
        elif re!={-1}:af.append(-1)
print("\n".join(map(str,af)))