codeforces 165E Compatible Numbers（位运算）【模板】

Two integers x and y are compatible, if the result of their bitwise "AND" equals zero, that is, a & b = 0. For example, numbers 90 (1011010₂) and 36 (100100₂) are compatible, as 1011010₂ & 100100₂ = 0₂, and numbers 3 (11₂) and 6 (110₂) are not compatible, as 11₂ & 110₂ = 10₂.

You are given an array of integers a₁, a₂, ..., a_n. Your task is to find the following for each array element: is this element compatible with some other element from the given array? If the answer to this question is positive, then you also should find any suitable element.

Input

The first line contains an integer n (1 ≤ n ≤ 10⁶) — the number of elements in the given array. The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 4·10⁶) — the elements of the given array. The numbers in the array can coincide.

Output

Print n integers ans_i. If a_i isn't compatible with any other element of the given array a₁, a₂, ..., a_n, then ans_i should be equal to -1. Otherwise ans_i is any such number, that a_i & ans_i = 0, and also ans_i occurs in the array a₁, a₂, ..., a_n.

Example

Input

2
90 36

Output

36 90

Input

4
3 6 3 6

Output

-1 -1 -1 -1

Input

5
10 6 9 8 2

Output

-1 8 2 2 8

 【题解】 比较委婉的DP题，给定一组数，判断里面的元素与其他元素与其他是否相溶，(即元素i与其他元素的亦或值都等于0，则输出-1，否则输出与其他任意元素的亦或值的值)， 我们可以每次把所有元素的亦或值存起来，处理完再输出。

 详解看代码注释

 【AC代码】

 

#include
#include
#include
#include
using namespace std;
const int N=1e6+5;
const int inf=(1<<22)-1;
int m,n;
int a[10000005],b[inf];

int main()
{
        scanf("%d",&m);
        for(int i=1;i<=m;++i)
        {
            scanf("%d",&a[i]);
            b[inf^a[i]]=a[i];//保存所有亦或值
        }
        for(int i=inf;i>=0;--i)//遍历一遍  
        {
            if(!b[i])//没有相应的亦或值
            {
                for(int j=0;j<22;++j)
                    if(b[i|(1<