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.

Source: codeforces | Difficulty: VERY_HARD
Tags: Graph algorithms

Data structures

Constructive algorithms


Solutions:
import sys
input = sys.stdin.readline
testcase = int(input())
for _ in range(testcase):
	(n, m, k) = map(int, input().split())
	ab = [list(map(int, input().split())) for i in range(m)]
	if k == 1:
		print(2)
		print(1)
		continue
	if k * (k - 1) > 2 * m:
		print(-1)
		continue
	graph = [set() for i in range(n + 1)]
	for (a, b) in ab:
		graph[a].add(b)
		graph[b].add(a)
	trash = set()
	stack = []
	for i in range(1, n + 1):
		if len(graph[i]) < k - 1:
			trash.add(i)
			stack.append(i)
	while stack:
		x = stack.pop()
		for y in graph[x]:
			if x in graph[y]:
				graph[y].remove(x)
				if len(graph[y]) < k - 1:
					stack.append(y)
					trash.add(y)
		graph[x] = set()
	if n - len(trash) < k:
		print(-1)
		continue
	ansflg = 0
	ccand = []
	for i in range(1, n + 1):
		if len(graph[i]) == k - 1:
			ccand.append(i)
	while ccand:
		x = ccand.pop()
		if len(graph[x]) < k - 1:
			trash.add(x)
			for y in graph[x]:
				if x in graph[y]:
					graph[y].remove(x)
					if len(graph[y]) == k - 1:
						ccand.append(y)
			graph[x] = set()
			continue
		lsg = list(graph[x])
		for i in range(len(lsg) - 1):
			for j in range(i + 1, len(lsg)):
				if lsg[j] not in graph[lsg[i]]:
					trash.add(x)
					for y in graph[x]:
						if x in graph[y]:
							graph[y].remove(x)
							if len(graph[y]) == k - 1:
								ccand.append(y)
					graph[x] = set()
					break
			else:
				continue
			break
		else:
			ansflg = 1
			break
	if ansflg:
		print(2)
		graph[x].add(x)
		print(*graph[x])
	elif n - len(trash) >= k + 1:
		ansls = []
		for i in range(1, n + 1):
			if i not in trash:
				ansls.append(i)
		print(1, len(ansls))
		print(*ansls)
	else:
		print(-1)


from __future__ import division, print_function
from heapq import *
import sys
t = list(map(int, sys.stdin.readline().strip().split()))[0]
af = []
for _ in range(t):
	(n, m, k) = list(map(int, sys.stdin.readline().strip().split()))
	ar = [set() for _ in range(n)]
	for i in range(m):
		(a, b) = list(map(int, sys.stdin.readline().strip().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)))


from collections import deque
import sys
input = sys.stdin.readline
for _ in range(int(input())):
	(n, m, k) = map(int, input().split())
	G = [set() for _ in range(n + 1)]
	deg = [0] * (n + 1)
	for _ in range(m):
		(i, j) = map(int, input().split())
		G[i].add(j)
		G[j].add(i)
		deg[i] += 1
		deg[j] += 1
	if k * (k - 1) > 2 * m:
		print(-1)
		continue
	q = deque()
	seen = [0] * (n + 1)
	for i in range(1, n + 1):
		if deg[i] < k:
			seen[i] = 1
			q.append(i)
	clique = []
	end = 0
	while q and (not end):
		i = q.popleft()
		seen[i] = 2
		if deg[i] == k - 1 and (not clique) and (k * (k - 1) // 2 <= m):
			clique = [i]
			for j in G[i]:
				if seen[j] <= 1:
					clique.append(j)
			ok = 1
			for x in clique:
				for y in clique:
					if x == y:
						break
					if y not in G[x]:
						ok = 0
						break
				if not ok:
					break
			if not ok:
				clique = []
			else:
				print(2)
				print(*clique)
				end = 1
				break
		for j in G[i]:
			deg[j] -= 1
			if deg[j] < k and (not seen[j]):
				seen[j] = 1
				q.append(j)
	if not end:
		subset = [i for i in range(1, n + 1) if not seen[i]]
		if subset:
			print('1 {}'.format(len(subset)))
			print(*subset)
		else:
			print(-1)


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)))