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 nn integer 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.

#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define rush() int T;cin>>T;while(T--)
#define go(a) while(cin>>a)
#define ms(a,b) memset(a,b,sizeof a)
#define E 1e-8
#define debug(a) cout<<"*"<<a<<"*"<<endl
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> Pair;
const int inf=0x3f3f3f3f;
const int N=100+5;int n,m,t;int i,j,k;ll a[N],ans[N];bool vis[N],flag=1;
void print()
{flag=0;for(int i=1;i<=n;i++){cout<<ans[i]<<" ";}cout<<endl;return;
}
void dfs(int now)
{if(now>n+1) return ;if(now==n+1) {print(); return ;}for(int i=1;i<=n;i++){if(vis[i]) continue;if( (ans[now-1]==a[i]*3) || (ans[now-1]*2==a[i])){vis[i]=1; ans[now]=a[i];dfs(now+1);vis[i]=0;}}
}
int32_t main()
{IOS; string s;while(cin>>n){for(i=1;i<=n;i++) cin>>a[i];ms(vis,0);for(int i=1;i<=n&&flag;i++){if(!vis[i]){vis[i]=1; ans[1]=a[i];dfs(2);vis[i]=0;}}}return 0;
}