Problem:
Score : 500 points

Problem Statement

There are N Snuke Cats numbered 1, 2, \ldots, N, where N is even.

Each Snuke Cat wears a red scarf, on which his favorite non-negative integer is written.

Recently, they learned the operation called xor (exclusive OR).

What is xor?

For n non-negative integers x_1, x_2, \ldots, x_n, their xor, x_1~\textrm{xor}~x_2~\textrm{xor}~\ldots~\textrm{xor}~x_n is defined as follows:

When x_1~\textrm{xor}~x_2~\textrm{xor}~\ldots~\textrm{xor}~x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even.

For example, 3~\textrm{xor}~5 = 6.

They wanted to use this operation quickly, so each of them calculated the xor of the integers written on their scarfs except his scarf.

We know that the xor calculated by Snuke Cat i, that is, the xor of the integers written on the scarfs except the scarf of Snuke Cat i is a_i.
Using this information, restore the integer written on the scarf of each Snuke Cat.

Constraints

All values in input are integers.

2 \leq N \leq 200000

N is even.

0 \leq a_i \leq 10^9

There exists a combination of integers on the scarfs that is consistent with the given information.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_N

Output

Print a line containing N integers separated with space.

The i-th of the integers from the left should represent the integer written on the scarf of Snuke Cat i.

If there are multiple possible solutions, you may print any of them.

Sample Input 1

4
20 11 9 24

Sample Output 1

26 5 7 22

5~\textrm{xor}~7~\textrm{xor}~22 = 20

26~\textrm{xor}~7~\textrm{xor}~22 = 11

26~\textrm{xor}~5~\textrm{xor}~22 = 9

26~\textrm{xor}~5~\textrm{xor}~7 = 24

Thus, this output is consistent with the given information.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/******************************************************************************
 
рассвет. 
 
*******************************************************************************/
#include <bits/stdc++.h>
typedef long long int lli;
typedef unsigned long long int ulli;
typedef long double ld;
#define rapgame ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb push_back
#define eb emplace_back
#define vi vector <int>
#define vlli vector <lli>
#define vulli vector <ulli>
#define vchar vector <char>
#define vstr vector <string>
#define vb vector <bool>
#define fst first
#define sec second
#define all(a) (a).begin(),(a).end()
#define srt(a) sort(a.begin(),a.end())
#define rev(a) reverse(a.begin(),a.end())
#define vpii vector <pair <int,int> >
#define pii pair <int,int> 
#define vplli vector <pair <lli,lli> >
#define vvi vector <vector <int> >
#define vvlli vector <vector <lli> >
#define vvpii vector <vector <pair <int,int>> >
#define fr(i,n) for (int i = 0; i < (n); ++i)
#define frbig(i,n) for (lli i = 0; i < (n); ++i)
#define fra(i,a,n) for (lli i = a; i < (n); ++i)
#define run(c,s) for(auto c:s) 
#define read(n) int n; cin >> n;
#define readchar(n) char n; cin >> n;
#define readstr(n) string n; cin >> n;
#define readarr(a,n) fr(i,n) cin >> a[i];
#define read2(x,y) int x,y; cin >> x >> y;
#define read2ll(x,y) long long int x,y; cin >> x >> y;
#define readll(n) lli n; cin >> n;
#define watch(x) cout << (#x) << " is " << (x) << endl
#define print_array(x) fr(i,x.size()) cout << x[i] << " ";
using namespace std;
 
const lli MOD=1e9+7;
const lli MAXN=1000000000000001;
const ld PI=acos(-1);
 
int main(){
    rapgame
    //freopen("unionday.in", "r", stdin);
    //freopen("unionday.out", "w", stdout);
    int t=1;
    //cin>>t;
    while(t--){
        readll(n);
        vlli a(n);
        bool bits[n][61],ans[n][61]; 
        fr(i,n){
            fr(j,61){
                bits[i][j]=ans[i][j]=0;
            }
        }
        fr(i,n){
            cin >> a[i];
            fr(j,61){
                if(a[i]>=(1ll<<(60-j))){
                    a[i]-=(1ll<<(60-j));
                    bits[i][60-j]=ans[i][60-j]=1;
                }
            }
        }
        // 10100 01011 01001 11000
        fr(i,61){
            int f=n-1;
            set <int> used;
            fr(j,n){
                bool k=bits[j][i],s=(((int)used.size())%2==1),del=(used.find(j)!=used.end());
                if(k!=s^del){ ans[f][i]^=1;used.insert(f);}
                --f;
            }
        }
        fr(i,n){
            lli b=0;
            fr(j,61){
                b+=(1ll<<j)*ans[i][j];
            }
            cout << b << " ";
        }
    }
    return 0;
}