链接：https://codeforces.com/contest/1300/problem/C

Anu has created her own function ff: f(x,y)=(x|y)−yf(x,y)=(x|y)−y where || denotes the bitwise OR operation. For example, f(11,6)=(11|6)−6=15−6=9f(11,6)=(11|6)−6=15−6=9. It can be proved that for any nonnegative numbers xx and yy value of f(x,y)f(x,y) is also nonnegative.

She would like to research more about this function and has created multiple problems for herself. But she isn't able to solve all of them and needs your help. Here is one of these problems.

A value of an array [a1,a2,…,an][a1,a2,…,an] is defined as f(f(…f(f(a1,a2),a3),…an−1),an)f(f(…f(f(a1,a2),a3),…an−1),an) (see notes). You are given an array with not necessarily distinct elements. How should you reorder its elements so that the value of the array is maximal possible?

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105).

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1090≤ai≤109). Elements of the array are not guaranteed to be different.

Output

Output nn integers, the reordering of the array with maximum value. If there are multiple answers, print any.

Examples

input

Copy

4
4 0 11 6

output

Copy

11 6 4 0

input

Copy

1
13

output

Copy

13

Note

In the first testcase, value of the array [11,6,4,0][11,6,4,0] is f(f(f(11,6),4),0)=f(f(9,4),0)=f(9,0)=9f(f(f(11,6),4),0)=f(f(9,4),0)=f(9,0)=9.

[11,4,0,6][11,4,0,6] is also a valid answer.

 

#include<bits/stdc++.h>
using namespace std;
long long n,t,s,k,max1;
struct node{
	int s[100];
	int x;
	int sum;
	int id;
}a[200005];
bool cmp(node q,node g)
{
	int k1=q.sum;
	int k2=g.sum;
	while(q.s[k1]==g.s[k2]&&k1>1&&k2>1)
	{
		k1--;
		k2--;
	}
	if(q.s[k1]!=g.s[k2])
	return q.s[k1]>g.s[k2];
	else
	return q.sum>g.sum;
}
map<long long,long long>m;
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i].x;
		k=a[i].x;
		int j=1;
		while(k)
		{
			if(k%2==1)
			{
				m[j]++;
			}
			k/=2;
			j++;
		}	
	}
	max1=0;
	for(int i=1;i<=n;i++)
	{
		k=a[i].x;
		int j=1,p=0;
		while(k)
		{
			if(k%2==1)
			{
				if(m[j]==1)
				{
					p++;
					a[i].s[p]=j;
				}
			}
			k/=2;
			j++;
		}
		a[i].sum=p;
	}
	sort(a+1,a+1+n,cmp);
	for(int i=1;i<=n;i++)
	{
		cout<<a[i].x<<" ";
	}
}