1407_B. Big Vova
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.

You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b. 

Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.

A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:

  * a is a prefix of b, but a ≠ b;
  * in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.

The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.

It is guaranteed that the sum of n over all test cases does not exceed 10^3.

Output

For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.

Example

Input


7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17


Output


5 2 
8 2 1 3 
9 3 8 
100 50 25 75 64 
42 
128 96 80 88 52 7 
17 2 4 8 16 

Note

In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].

In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.

In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
```python
import sys,bisect,string,math,time,functools,random,fractions
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
rep=range;R=range
def Golf():n,*t=map(int,open(0).read().split())
def I():return int(input())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def MI():return map(int,input().split())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RA():return map(int,open(0).read().split())
def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
def RI(a=1,b=10):return random.randint(a,b)
def Rtest(T):
    case,err=0,0
    for i in range(T):
        inp=INP()
        a1,ls=naive(*inp)
        a2=solve(*inp)
        if a1!=a2:
            print((a1,a2),inp)
            err+=1
        case+=1
    print('Tested',case,'case with',err,'errors')
def GI(V,E,ls=None,Directed=False,index=1):
    org_inp=[];g=[[] for i in range(V)]
    FromStdin=True if ls==None else False
    for i in range(E):
        if FromStdin:
            inp=LI()
            org_inp.append(inp)
        else:
            inp=ls[i]
        if len(inp)==2:
            a,b=inp;c=1
        else:
            a,b,c=inp
        if index==1:a-=1;b-=1
        aa=(a,c);bb=(b,c);g[a].append(bb)
        if not Directed:g[b].append(aa)
    return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
    #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
    mp=[boundary]*(w+2);found={}
    for i in R(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
    mp+=[boundary]*(w+2)
    return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def accum(ls):
    rt=[0]
    for i in ls:rt+=[rt[-1]+i]
    return rt
def bit_combination(n,base=2):
    rt=[]
    for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
    return rt
def gcd(x,y):
    if y==0:return x
    if x%y==0:return y
    while x%y!=0:x,y=y,x%y
    return y
def YN(x):print(['NO','YES'][x])
def Yn(x):print(['No','Yes'][x])
def show(*inp,end='\n'):
    if show_flg:print(*inp,end=end)

mo=10**9+7
#mo=998244353
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
l_alp=string.ascii_lowercase
#sys.setrecursionlimit(10**9)
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()


## return prime factors of N as dictionary {prime p:power of p}
## within 2 sec for N = 2*10**20+7
def primeFactor(N):
    i,n=2,N
    ret={}
    d,sq=2,99
    while i<=sq:
        k=0
        while n%i==0:
            n,k,ret[i]=n//i,k+1,k+1
        if k>0 or i==97:
            sq=int(n**(1/2)+0.5)
        if i<4:
            i=i*2-1
        else:
            i,d=i+d,d^6
    if n>1:
        ret[n]=1
    return ret

## return divisors of n as list
def divisor(n):
    div=[1]
    for i,j in primeFactor(n).items():
        div=[(i**k)*d for d in div for k in range(j+1)]
    return div

## return the list of prime numbers in [2,N], using eratosthenes sieve
## around 800 ms for N = 10**6  by PyPy3 (7.3.0) @ AtCoder
def PrimeNumSet(N):
    M=int(N**0.5)
    seachList=[i for i in range(2,N+1)]
    primes=[]
    while seachList:
        if seachList[0]>M:
            break
        primes.append(seachList[0])
        tmp=seachList[0]
        seachList=[i for i in seachList if i%tmp!=0]
    return primes+seachList


## retrun LCM of numbers in list b
## within 2sec for no of B = 10*5  and  Bi < 10**6
def LCM(b,mo=10**9+7):
    prs=PrimeNumSet(max(b))
    M=dict(zip(prs,[0]*len(prs)))
    for i in b:
        dc=primeFactor(i)
        for j,k in dc.items():
            M[j]=max(M[j],k)
    
    r=1
    for j,k in M.items():
        if k!=0:
            r*=pow(j,k,mo)
            r%=mo
    return r

show_flg=False
show_flg=True

ans=0


for _ in range(I()):
    n=I()
    a=LI()
    a=sorted(a)
    m=a.pop()
    ans=[m]
    v=[0]*(n-1)
    ko=[[]for i in range(m+1)]
    for i in range(n-1):
        c=a[i]
        for j in divisor(c):
            ko[j]+=i,
    #show(ko)
    #for i in range(n):
    ans=[]
    g=m
    while g>1:
        dm=sorted(divisor(g))[::-1]
        tmp=[]
        for d in dm:
            if ko[d]:
                for x in ko[d]:
                    if v[x]==0:
                        v[x]=1
                        tmp+=x,
                if tmp:
                    g=gcd(g,a[tmp[0]])
                    ans+=tmp
                    break
        if sum(v)==n-1:
            break
    t=[m]+[a[i] for i in ans]
    print(*t)
    #show(a)



```