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

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⋅1018

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

6
4 8 6 3 12 9

9 3 6 12 4 8 

4
42 28 84 126

126 42 84 28 

2
1000000000000000000 3000000000000000000

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.

題意：

給你一個含有N個數的數組，讓你重新定義他們的順序，使之a[i] 和a[i-1] 只有兩種關系

1. a[i]=a[i-1]/3

2. a[i]=a[ i-1 ] * 2

思路：

暴力DFS竟然騙過，難以置信。

正解的思路是：

我們定義一個數x的cnt3是它唯一分解後3的次冪數。

因為只有/3的操作，所以整個數組中cnt3一定是單調不遞增的，

而相同的cnt3中的數，關系一定是*2的關系，那麽一定是單調遞增的排序。

那麽我們只需要把數都壓入到一個pair中，first 是這個數的cnt3，second是這個數，然後進行默認的pair排序。

默認的pair排序是，先以first遞增排序，相同的first的以second遞增排序。

那麽我們把每一個數的cnt3取為負值，即-cnt3。

那麽就直接排序就輸出結果了。

細節見代碼：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int count3(LL x){
  int ret=0;
  while(x % 3 == 0){
    ret++;
    x /= 3;
  }
  return ret;
}
int n;
vector<pair<int,LL>> v;
int main(){
  cin>>n;
  v.resize(n);
  for(int i=0; i<n; i++){
    cin>>v[i].second;
    v[i].first=-count3(v[i].second);
  }
  sort(v.begin(), v.end());
  for(int i=0;i<n;i++)
  {
      printf("%d %lld\n",v[i].first, v[i].second);
  }
  for(int i=0; i<n; i++)
    printf("%lld%c", v[i].second, " \n"[i + 1 == n]);
}

Divide by three, multiply by two CodeForces - 977D （思維排序）