CF722D. Generating Sets[贪心 STL]

You are given a set Y of n distinct positive integers y1, y2, ..., yn.

Set X of n distinct positive integers x1, x2, ..., xn is said to generate set Y if one can transform X to Y by applying some number of the following two operation to integers in X:

Take any integer xi and multiply it by two, i.e. replace xi with 2·xi.
Take any integer xi, multiply it by two and add one, i.e. replace xi with 2·xi + 1.

Note that integers in X are not required to be distinct after each operation.

Two sets of distinct integers X and Y are equal if they are equal as sets. In other words, if we write elements of the sets in the array in the increasing order, these arrays would be equal.

Note, that any set of integers (or its permutation) generates itself.

You are given a set Y and have to find a set X that generates Y and the maximum element of X is mininum possible.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 50 000) — the number of elements in Y.

The second line contains n integers y1, ..., yn (1 ≤ yi ≤ 109), that are guaranteed to be distinct.

Output

Print n integers — set of distinct integers that generate Y and the maximum element of which is minimum possible. If there are several such sets, print any of them.

Examples

input

5
1 2 3 4 5

output

4 5 2 3 1

input

6
15 14 3 13 1 12

output

12 13 14 7 3 1

input

6
9 7 13 17 5 11

output

4 5 2 6 3 1

题意：找一个集合X，使得集合y中的元素都可以由集合x进行*2或*2+1得到，可以操作由操作得来的元素，求最大元素最小的集合x

还有30分钟开始做这道题 
想到贪心方法了每次取最大x然后x/2，如果当前没有x/2就加入，否则让x/2再/2
然而当时读错题意了把新得到的加入了一个新集合里

WARN：是正整数，没有0

//
//  main.cpp
//  d
//
//  Created by Candy on 10/1/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
using namespace std;
const int N=5e4+5;
inline int read(){char c=getchar();int x=0,f=1;while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}return x;
}
int n,y[N],cnt=0;
set<int> s;
set<int>::iterator it;
int main(int argc, const char * argv[]) {n=read();for(int i=1;i<=n;i++) {y[i]=read();s.insert(y[i]);}while(true){it=s.end(); it--;int x=*it;while(s.find(x)!=s.end()&&x) x>>=1;if(x==0) break;else{s.erase(it);s.insert(x);}}for(it=s.begin();it!=s.end();it++) printf("%d ",*it);return 0;
}