题干：

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 

Note

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

题目大意：

给定n个数，找到他的一个重新排列，使得第一个数开始通过除3或乘2，可以得到这个序列。（n<=100. a[i]<=3e18）

解题报告：

设cnt3是一个数中3的因子个数。

因为只有/3的操作，所以排完的整个数组中的数cnt3一定是单调不递增的，

而相同的cnt3中的数，关系一定是*2的关系，那么一定是单调递增的排序。

所以先按cnt3降序排序，cnt3相同的升序排序，直接输出即可。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n;
struct Node {ll a,b;bool operator < (const Node & o) const {if(b != o.b) return b > o.b;else return a < o.a;}
} R[MAX];
int main()
{cin>>n;for(int i = 1; i<=n; i++) {scanf("%lld",&R[i].a);ll x = R[i].a;while(x%3 == 0) x/=3,R[i].b++;}sort(R+1,R+n+1);for(int i = 1; i<=n; i++) printf("%lld ",R[i].a);return 0 ;
}