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.

?

題意：

給你一個含有N個數(shù)的數(shù)組，讓你重新定義他們的順序，使之a(chǎn)[i] 和a[i-1] 只有兩種關(guān)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(shù)x的cnt3是它唯一分解后3的次冪數(shù)。

因為只有/3的操作，所以整個數(shù)組中cnt3一定是單調(diào)不遞增的，

而相同的cnt3中的數(shù)，關(guān)系一定是*2的關(guān)系，那么一定是單調(diào)遞增的排序。

那么我們只需要把數(shù)都壓入到一個pair中，first 是這個數(shù)的cnt3，second是這個數(shù)，然后進行默認的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。

那么我們把每一個數(shù)的cnt3取為負值，即-cnt3。

那么就直接排序就輸出結(jié)果了。

細節(jié)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]); }

?

轉(zhuǎn)載于:https://www.cnblogs.com/qieqiemin/p/10664418.html

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