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Polycarp likes to play with numbers. He takes some integer number x, writes it down on the board, and then performs with it n−1 operations of the two kinds:

• divide the number x by 3 (x must be divisible by 3);
• multiply the number x by 2.

After each operation, Polycarp writes down the result on the board and replaces x by the result. So there will be n numbers on the board after all.

You are given a sequence of length n — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.

Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.

It is guaranteed that the answer exists.

Input

The first line of the input contatins an integer number n (2≤n≤100) — the number of the elements in the sequence. The second line of the input contains n integer numbers a1,a2,…,an (1≤ai≤3⋅10^18) — rearranged (reordered) sequence that Polycarp can wrote down on the board.

Output

Print n integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.

It is guaranteed that the answer exists.

Examples

Input

64 8 6 3 12 9

Output

9 3 6 12 4 8

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Input

442 28 84 126

Output

126 42 84 28

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Input

21000000000000000000 3000000000000000000

Output

3000000000000000000 1000000000000000000

Note

In the first example the given sequence can be rearranged in the following way: [9,3,6,12,4,8]. It can match possible Polycarp's game which started with x=9.

题意：

给出一个整数n，然后给乱序的n个数，这个n个数是由一个数通过/3或者*2得到的另外n-1个数，要求我们将正确的顺序输出。

通过题目我们可以知道，要从数a得到b数b，一定是通过a/3=b 或者 a*2=b ，所以我们可以找到n个数中的关系，然后使用拓扑排序得到原来的顺序。 

代码：

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        #include
        
         using namespace std;typedef long long ll;const int maxn=117;int n;vector
         
           path[maxn]; //存储数之间的关系ll num[maxn];int rudu[maxn]; ll res[maxn]; //存放结果int cnt=0; int main(){ memset(rudu,0,sizeof(rudu)); scanf("%d",&n); for(int i=0;i
         
        
       
      
     
    
   

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