D. Divide by three, multiply by two(要什么脑子，暴力吖)

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 $n$ ( $2 \leq n \leq 100$ ) — the number of the elements in the sequence. The second line of the input contains $n$ integer numbers $a_{1}, a_{2}, \dots, a_{n}$ ( $1 \leq a_{i} \leq 3 \cdot 10^{18}$ ) — 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 
          
            Copy 
          
6
4 8 6 3 12 9

           output 
          
            Copy 
          
9 3 6 12 4 8 

           input 
          
            Copy 
          
4
42 28 84 126

           output 
          
            Copy 
          
126 42 84 28 

           input 
          
            Copy 
          
2
1000000000000000000 3000000000000000000

           output 
          
            Copy 
          
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 $x = 9$ .

看题目，给你一个序列，重新排序使得新数列满足以下条件

后一个数要么是前一个数的三分之一，要么是两倍。完。。。

仔细看一遍题，保证一定有解，并且，解唯一（没说多组答案）

然后就好玩了，随便从原数列中抽一个数，向后向前延伸即可，

本来是要dfs的，后来一想，题目这么氵，那就直接写好了

时间复杂度大概emm直接就是O（n）吧

ac代码如下

#include<bits/stdc++.h>
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<string.h>
#include <queue>
using namespace std;
typedef long long ll;
ll cnt[105];
ll num[300];
int n;
map<ll,int> P;
int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>cnt[i],P[cnt[i]]++;

    num[105]=cnt[0];

    for(int i=105;;i++)
    {
        ll c=num[i];
        if(c%3==0 && P[c/3])
        {
            P[c/3]--;
            num[i+1]=c/3;
        }
        else if(P[c*2])
        {
            P[c*2]--;
            num[i+1]=c*2;
        }
        if(!num[i+1])
            break;
    }

    for(int i=105;;i--)
    {
        ll c=num[i];
        if(P[c*3])
        {
            P[c*3]--;
            num[i-1]=c*3;
        }
        else if(c%2==0 && P[c/2])
        {
            P[c/2]--;
            num[i-1]=c/2;
        }
        if(!num[i-1])
            break;
    }

    int i=0;

    while(!num[i])i++;

    printf("%lld",num[i++]);

    while(num[i])printf(" %lld",num[i++]);

    printf("\n");


    return 0;
}

D. Divide by three, multiply by two(要什么脑子，暴力吖)

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