Divide by three, multiply by two

Polycarp likes to play with numbers. He takes some integer number $\mathrm{xx,\; writes\; it\; down\; on\; the\; board,\; and\; then\; performs\; with\; it\; n-1n-1\; operations\; of\; the\; two\; kinds:}$

• divide the number $\mathrm{xx\; by\; 33\; (xx\; must\; be\; divisible\; by\; 33);}$
• multiply the number $\mathrm{xx\; by\; 22.}$

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

You are given a sequence of length $\mathrm{nn\; —\; the\; numbers\; that\; Polycarp\; wrote\; down.\; This\; sequence\; is\; given\; in\; arbitrary\; order,\; i.e.\; the\; order\; of\; the\; sequence\; can\; mismatch\; the\; order\; of\; the\; numbers\; written\; on\; the\; board.}$

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 $\mathrm{nn\; (2\le n\le 1002\le n\le 100)\; —\; the\; number\; of\; the\; elements\; in\; the\; sequence.\; The\; second\; line\; of\; the\; input\; contains\; nn\; integer\; numbers\; a1,a2,\dots ,ana1,a2,\dots ,an\; (1\le ai\le 3\cdot 10181\le ai\le 3\cdot 1018)\; —\; rearranged\; (reordered)\; sequence\; that\; Polycarp\; can\; wrote\; down\; on\; the\; board.}$

Output

Print $\mathrm{nn\; integer\; numbers\; —\; rearranged\; (reordered)\; input\; sequence\; that\; can\; be\; the\; sequence\; that\; Polycarp\; could\; write\; down\; on\; the\; board.}$

It is guaranteed that the answer exists.

Examples

Input

Copy

6
4 8 6 3 12 9

Output

Copy

9 3 6 12 4 8 

Input

Copy

4
42 28 84 126

Output

Copy

126 42 84 28 

Input

Copy

2
1000000000000000000 3000000000000000000

Output

Copy

3000000000000000000 1000000000000000000 

Note

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

$题解：找每个点的后继$

#pragma warning(disable:4996)
#include<cmath>
#include<queue>
#include<map>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long

map<ll, int> p;
vector<ll> v;

const int maxn = 105;

ll n, m;
ll a[maxn];

bool DFS(ll num, int dep) {
    bool res = false;
    if (dep == n) {
        v.push_back(num);
        return true;
    }
    if (num % 3 == 0 && p[num / 3]) {
        if (DFS(num / 3, dep + 1)) {
            v.push_back(num);
            res = true;
        }
    }
    if (num * 2 <= m && p[num * 2]) {
        if (DFS(num * 2, dep + 1)) {
            v.push_back(num);
            res = true;
        }  
    }
    return res;
}

int main()
{
    while (cin >> n) {
        m = 0;
        p.clear();
        v.clear();
        for (int i = 1; i <= n; i++) {
            cin >> a[i];
            p[a[i]]++;
            m = max(m, a[i]);
        }
        for (int i = 1; i <= n; i++) {
            if (DFS(a[i], 1)) break;
            else v.clear();
        }
        for (int i = n - 1; i >= 0; i--) cout << v[i] << endl;
    }
    return 0;
}

 代码2：

#pragma warning(disable:4996)
#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long

map<ll, int> p;

const int maxn = 105;

ll n, m;
ll a[maxn], ans[maxn];
bool flag;

void DFS(ll num, int dep) {
    ans[dep] = num;
    if (dep == n) {
        flag = true;
        return;
    }
    if (num % 3 == 0 && p[num / 3]) DFS(num / 3, dep + 1);
    if (num * 2 <= m && p[num * 2]) DFS(num * 2, dep + 1);
    return;
}

int main()
{
    while (cin >> n) {
        m = 0;
        p.clear();
        for (int i = 1; i <= n; i++) {
            cin >> a[i];
            p[a[i]]++;
            m = max(m, a[i]);
        }
        for (int i = 1; i <= n; i++) {
            flag = false;
            DFS(a[i], 1);
            if (flag) break;
        }
        for (int i = 1; i <= n; i++) {
            cout << ans[i] << endl;
        }
    }
    return 0;
}