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:
''' CODED WITH LOVE BY SATYAM KUMAR '''

from sys import stdin, stdout
import heapq
import cProfile, math
from collections import Counter, defaultdict, deque
from bisect import bisect_left, bisect, bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
import operator as op
from functools import reduce
import sys

sys.setrecursionlimit(10 ** 6)  # max depth of recursion
threading.stack_size(2 ** 27)  # new thread will get stack of such size
fac_warm_up = False
printHeap = str()
memory_constrained = False
P = 10 ** 9 + 7


class MergeFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n
        self.num_sets = n
        self.lista = [[_] for _ in range(n)]

    def find(self, a):
        to_update = []
        while a != self.parent[a]:
            to_update.append(a)
            a = self.parent[a]
        for b in to_update:
            self.parent[b] = a
        return self.parent[a]

    def merge(self, a, b):
        a = self.find(a)
        b = self.find(b)
        if a == b:
            return
        if self.size[a] < self.size[b]:
            a, b = b, a
        self.num_sets -= 1
        self.parent[b] = a
        self.size[a] += self.size[b]
        self.lista[a] += self.lista[b]

    def set_size(self, a):
        return self.size[self.find(a)]

    def __len__(self):
        return self.num_sets


def display(string_to_print):
    stdout.write(str(string_to_print) + "\n")


def prime_factors(n):  # n**0.5 complex
    factors = dict()
    for i in range(2, math.ceil(math.sqrt(n)) + 1):
        while n % i == 0:
            if i in factors:
                factors[i] += 1
            else:
                factors[i] = 1
            n = n // i
    if n > 2:
        factors[n] = 1
    return (factors)


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


def fibonacci_modP(n, MOD):
    if n < 2: return 1
    return (cached_fn(fibonacci_modP, (n + 1) // 2, MOD) * cached_fn(fibonacci_modP, n // 2, MOD) + cached_fn(
        fibonacci_modP, (n - 1) // 2, MOD) * cached_fn(fibonacci_modP, (n - 2) // 2, MOD)) % MOD


def factorial_modP_Wilson(n, p):
    if (p <= n):
        return 0
    res = (p - 1)
    for i in range(n + 1, p):
        res = (res * cached_fn(InverseEuler, i, p)) % p
    return res


def binary(n, digits=20):
    b = bin(n)[2:]
    b = '0' * (digits - len(b)) + b
    return b


def is_prime(n):
    """Returns True if n is prime."""
    if n < 4:
        return True
    if n % 2 == 0:
        return False
    if n % 3 == 0:
        return False
    i = 5
    w = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += w
        w = 6 - w
    return True


def generate_primes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    return prime


factorial_modP = []


def warm_up_fac(MOD):
    global factorial_modP, fac_warm_up
    if fac_warm_up: return
    factorial_modP = [1 for _ in range(fac_warm_up_size + 1)]
    for i in range(2, fac_warm_up_size):
        factorial_modP[i] = (factorial_modP[i - 1] * i) % MOD
    fac_warm_up = True


def InverseEuler(n, MOD):
    return pow(n, MOD - 2, MOD)


def nCr(n, r, MOD):
    global fac_warm_up, factorial_modP
    if not fac_warm_up:
        warm_up_fac(MOD)
        fac_warm_up = True
    return (factorial_modP[n] * (
                (pow(factorial_modP[r], MOD - 2, MOD) * pow(factorial_modP[n - r], MOD - 2, MOD)) % MOD)) % MOD


def test_print(*args):
    if testingMode:
        print(args)


def display_list(list1, sep=" "):
    stdout.write(sep.join(map(str, list1)) + "\n")


def display_2D_list(li):
    for i in li:
        print(i)


def prefix_sum(li):
    sm = 0
    res = []
    for i in li:
        sm += i
        res.append(sm)
    return res


def get_int():
    return int(stdin.readline().strip())


def get_tuple():
    return map(int, stdin.readline().split())


def get_list():
    return list(map(int, stdin.readline().split()))


memory = dict()


def clear_cache():
    global memory
    memory = dict()


def cached_fn(fn, *args):
    global memory
    if args in memory:
        return memory[args]
    else:
        result = fn(*args)
        memory[args] = result
        return result


def ncr(n, r):
    return math.factorial(n) / (math.factorial(n - r) * math.factorial(r))


def binary_search(i, li):
    fn = lambda x: li[x] - x // i
    x = -1
    b = len(li)
    while b >= 1:
        while b + x < len(li) and fn(b + x) > 0:  # Change this condition 2 to whatever you like
            x += b
        b = b // 2
    return x


# -------------------------------------------------------------- MAIN PROGRAM


TestCases = True
fac_warm_up_size = 10 ** 5 + 100
optimise_for_recursion = False  # Can not be used clubbed with TestCases WHen using recursive functions, use Python 3


def main():
    n, m = get_tuple()
    adj = [[] for _ in range(n)]
    visited = [False for _ in range(n)]
    for _ in range(m):
        x, y = get_tuple()
        adj[x-1].append(y-1)
        adj[y-1].append(x-1)
    queue = [0]
    mk = 1
    res = [[],[]]
    visited[0] = True
    while len(queue)>0:
        mk =mk^1
        temp_q = []
        for vertex in queue:

            res[mk].append(vertex+1)
            for j in adj[vertex]:
                if not visited[j]:
                    temp_q.append(j)
                    visited[j] = True
        queue = temp_q
    #print(res)
    print(min(len(res[0]), len(res[1])))
    display_list(res[0]) if len(res[0])<len(res[1]) else display_list(res[1])

# --------------------------------------------------------------------- END=


if TestCases:
    for i in range(get_int()):
        main()
else:
    main() if not optimise_for_recursion else threading.Thread(target=main).start()