Problem: You are given an undirected unweighted connected graph consisting of n vertices and m edges. It is guaranteed that there are no self-loops or multiple edges in the given graph.

Your task is to choose at most ⌊n/2⌋ vertices in this graph so each unchosen vertex is adjacent (in other words, connected by an edge) to at least one of chosen vertices.

It is guaranteed that the answer exists. If there are multiple answers, you can print any.

You will be given multiple independent queries to answer.

Input

The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of queries.

Then t queries follow.

The first line of each query contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5, n - 1 ≤ m ≤ min(2 ⋅ 10^5, (n(n-1))/(2))) — the number of vertices and the number of edges, respectively.

The following m lines denote edges: edge i is represented by a pair of integers v_i, u_i (1 ≤ v_i, u_i ≤ n, u_i ≠ v_i), which are the indices of vertices connected by the edge.

There are no self-loops or multiple edges in the given graph, i. e. for each pair (v_i, u_i) there are no other pairs (v_i, u_i) or (u_i, v_i) in the list of edges, and for each pair (v_i, u_i) the condition v_i ≠ u_i is satisfied. It is guaranteed that the given graph is connected.

It is guaranteed that ∑ m ≤ 2 ⋅ 10^5 over all queries.

Output

For each query print two lines.

In the first line print k (1 ≤ ⌊n/2⌋) — the number of chosen vertices.

In the second line print k distinct integers c_1, c_2, ..., c_k in any order, where c_i is the index of the i-th chosen vertex.

It is guaranteed that the answer exists. If there are multiple answers, you can print any.

Example

Input


2
4 6
1 2
1 3
1 4
2 3
2 4
3 4
6 8
2 5
5 4
4 3
4 1
1 3
2 3
2 6
5 6


Output


2
1 3
3
4 3 6

Note

In the first query any vertex or any pair of vertices will suffice.

<image>

Note that you don't have to minimize the number of chosen vertices. In the second query two vertices can be enough (vertices 2 and 4) but three is also ok.

<image>

Solution (in PYTHON3):

Solution:
import sys
input = sys.stdin.readline


def factorial(x, m):
	val = 1
	while x > 0:
		val = (val * x) % m
		x -= 1
	return val


def fact(x):
	val = 1
	while x > 0:
		val *= x
		x -= 1
	return val


# swap_array function
def swaparr(arr, a, b):
	temp = arr[a]
	arr[a] = arr[b]
	arr[b] = temp


def gcd(a, b):
	if b == 0:
		return a
	return gcd(b, a % b)

def nCr(n, k):
	if k > n:
		return 0
	if (k > n - k):
		k = n - k
	res = 1
	for i in range(k):
		res = res * (n - i)
		res = res / (i + 1)
	return int(res)

## prime factorization
def primefs(n):
	primes = {}
	while (n % 2 == 0 and n > 0):
		primes[2] = primes.get(2, 0) + 1
		n = n // 2
	for i in range(3, int(n ** 0.5) + 2, 2):
		while (n % i == 0 and n > 0):
			primes[i] = primes.get(i, 0) + 1
			n = n // i
	if n > 2:
		primes[n] = primes.get(n, 0) + 1
	## prime factoriazation of n is stored in dictionary
	## primes and can be accesed. O(sqrt n)
	return primes


## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
	res = 1
	x = x % p
	if (x == 0):
		return 0
	while (y > 0):
		if ((y & 1) == 1):
			res = (res * x) % p
		y = y >> 1
		x = (x * x) % p
	return res

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
	prime = [True for i in range(n + 1)]
	prime[0], prime[1] = False, False
	p = 2
	while (p * p <= n):
		if (prime[p] == True):
			for i in range(p * p, n + 1, p):
				prime[i] = False
		p += 1
	return prime

def is_prime(n):
	if n == 0:
		return False
	if n == 1:
		return True
	for i in range(2, int(n ** (1 / 2)) + 1):
		if not n % i:
			return False

	return True


#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
import math
def spf_sieve():
	spf[1] = 1
	for i in range(2, MAXN):
		spf[i] = i
	for i in range(4, MAXN, 2):
		spf[i] = 2
	for i in range(3, math.ceil(MAXN ** 0.5), 2):
		if spf[i] == i:
			for j in range(i * i, MAXN, i):
				if spf[j] == j:
					spf[j] = i

spf = [0 for i in range(MAXN)]

# spf_sieve();
def factoriazation(x):
	res = []
	for i in range(2, int(x ** 0.5) + 1):
		while x % i == 0:
			res.append(i)
			x //= i
	if x != 1:
		res.append(x)
	return res

def factors(n):
	res = []
	for i in range(1, int(n ** 0.5) + 1):
		if n % i == 0:
			res.append(i)
			res.append(n // i)
	return list(set(res))

def int_array():
	return list(map(int, input().strip().split()))

def float_array():
	return list(map(float, input().strip().split()))


def str_array():
	return input().strip().split()

# defining a couple constants
MOD = int(1e9) + 7
CMOD = 998244353
INF = float('inf')
NINF = -float('inf')

################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

from itertools import permutations
import math
from bisect import bisect_left
#for _ in range(int(input())):
from collections import Counter
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
import heapq
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
# n = int(input())
# ls = list(map(int, input().split()))
# n, k = map(int, input().split())
# n =int(input())
# arr=[(i,x) for i,x in enum]
# arr.sort(key=lambda x:x[0])
# e=list(map(int, input().split()))
from collections import Counter
# print("\n".join(ls))
# print(os.path.commonprefix(ls[0:2]))
# n=int(input())
from bisect import bisect_right
# d=sorted(d,key=lambda x:(len(d[x]),-x))  d=dictionary     d={x:set() for x in arr}
# n=int(input())
# n,m,k= map(int, input().split())
import heapq
# for _ in range(int(input())):
# n,k=map(int, input().split())
#for _ in range(int(input())):
#n = int(input())
# code here ;))

#n,k=map(int, input().split())
#arr = list(map(int, input().split()))
#ls= list(map(int, input().split()))
import math
#n = int(input())
#n,k=map(int, input().split())
#for _ in range(int(input())):
#n = int(input())
#n,l=map(int,input().split())
#for _ in range(int(input())):
def divisors_seive(n):
	divisors=[10**6+6]
	for i in range(1,n+1):
		for j in range(i,n+1):
			divisors[j]+=1

def dfs(s,par,vis,rank):
	vis[s]=1
	global res
	res=1
	rank[s]=1
	f=0
	for i in g[s]:
		f=1
		if par==i:
			continue
		rank[s]+=dfs(i,s)
	#if rank[s]%2==0 and s!=1:
		#res+=1
	#if f==0:
	return res
#n=int(input())
#n=int(input())
import bisect
from collections import deque
n,m=map(int,input().split())
vis=[0]*(n+1)
q=[]
g=[[] for  i in range(n)]
ind=[0]*n
for i in range(m):
	u,v= map(int, input().split())
	g[u-1].append(v-1)
	g[v-1].append(u-1)
	ind[u-1]+=1
	ind[v-1]+=1

node=ind.index(max(ind))
q.append(node)
ans=[]
while q:
	s=q.pop()
	vis[s]=1
	for i in g[s]:
		if vis[i]==0:
			vis[i]=1
			q.append(i)
			ans.append([s+1,i+1])
for i in ans:
	print(*i)