2018.5.10训练题

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

• divide the number xx by 33 (xx must be divisible by 33);
• multiply the number xx by 22.

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

You are given a sequence of length nn — 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 nn (2≤n≤1002≤n≤100) — the number of the elements in the sequence. The second line of the input contains nninteger numbers a1,a2,…,ana1,a2,…,an (1≤ai≤3⋅10181≤ai≤3⋅1018) — rearranged (reordered) sequence that Polycarp can wrote down on the board.

Output

Print nn 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

6
4 8 6 3 12 9

Output

9 3 6 12 4 8 

Input

4
42 28 84 126

Output

126 42 84 28 

Input

2
1000000000000000000 3000000000000000000

Output

3000000000000000000 1000000000000000000

大体意思即给你一串数组，这组数是乱排的，要求你排序并且满足如下条件：前面的数是后面的数的三倍，如果不满足这个条件，则后面的数应该是前面的数的两倍。

思路：这个是从网上找的思路，首先分析一下，假设将所有数中能够整除3的排序，多的在前面，少的在后面，剩下的因为i是递增序列，所以后面的按照递增排序，用sort即可、

代码如下：

#include<bits/stdc++.h>
using namespace std;
long long int a[120];
long long int f(long long int temp)
{
long long int num=0;
while(temp%3==0)
{
           temp=temp/3;
           num++;
}
return num;
}
bool cmp(long long int t1,long long int t2)
{
long long int x=f(t1),y=f(t2);
if(x!=y)
           return x>y;


           return t1<t2;
}
int main()
{
long long int n;
cin>>n;
for(int i=1;i<=n;i++)
{
           cin>>a[i];
}
sort(a+1,a+n+1,cmp);
cout<<a[1];
for(int i=2;i<=n;i++)
           cout<<" "<<a[i];
cout<<endl;
return 0;
}