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: Accepted)

#include <iostream>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <cctype>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <stdlib.h>
#include <tuple>
#include <iomanip>
#include <iterator>
#include <stdio.h>
#include <fstream>
#include <time.h>
using namespace std;
const double PI = 3.14159265358979323846;
const long long int MOD = 1000000000 + 7;

struct UnionFind {
    vector<int> parent;

    UnionFind(int N) : parent(N) {
        for (int i = 0; i < N; i++) {
            parent[i] = -1;
        }
    }

    int root(int i) {
        if (parent[i] < 0) {
            return i;
        }
        return (parent[i] = root(parent[i]));
    }

    bool unite(int from, int to) {
        int rx = root(from);
        int ry = root(to);
        if (rx != ry) {
            parent[ry] += parent[rx];
            parent[rx] = ry;

            return true;
        }
        else {
            return false;
        }
    }

    bool same(int x, int y) {
        return root(x) == root(y);
    }

    int treeSize(int x) {


        return -parent[root(x)];
    }
};

long long int modpow(long long int base, long long int pow, long long int mod) {
    if (pow == 1) {
        return base;
    }
    else if (pow == 0) {
        return 1;
    }

    if (pow % 2 == 0) {
        auto temp = modpow(base, pow / 2, mod);
        return (temp * temp) % mod;
    }
    else {
        return (base * modpow(base, pow - 1, mod)) % mod;
    }
}

long long int moddiv(long long int X, long long int Y, long long int mod) {
    auto fermatDiv = modpow(Y, mod - 2, mod);

    return (X * fermatDiv) % mod;
}

long long modCombination(long long left, long long right, long long int mod) {
    long long answer = 1;
    if (left > right) {
        for (long long i = 0; i < right; i++) {
            answer = (answer * (left - i)) % mod;
            answer = moddiv(answer, (i + 1), mod);
        }
    }
    return answer;
}

bool IsPrime(long long N) {
    if (N == 1) {
        return false;
    }
    for (long long i = 2; i * i <= N; i++) {
        if (N % i == 0) {
            return false;
        }
    }


    return true;
}



vector<pair<long long, long long> > prime_factorize(long long N) {
    vector<pair<long long, long long> > res;
    for (long long a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        long long ex = 0; // 指数

        // 割れる限り割り続ける
        while (N % a == 0) {
            ++ex;
            N /= a;
        }

        // その結果を push
        res.push_back({ a, ex });
    }

    // 最後に残った数について
    if (N != 1) res.push_back({ N, 1 });
    return res;
}

long long euclid(long long a, long long b) {
    if (a > b) {
        long long temp = b;
        b = a;
        a = temp;
    }
    if (b % a == 0) {
        return a;
    }
    else {
        return euclid(a, b - a);
    }
}

long long gcd(long long a, long long b) {
    if (b > a) {
        long long temp = b;
        b = a;
        a = temp;
    }
    //cout << "a:" << a << "b:" << b << endl;
    long long c = a % b;
    if (c == 0) {
        return b;
    }
    else {
        return gcd(b, c);
    }
}


struct State {
    long long town;
    long long money;
    long long time;

    State(long long _town, long long _money, long long _time) {
        town = _town;
        money = _money;
        time = _time;
    }

    bool operator<(const State& s) const {
        return this->time > s.time;
    }
};

struct Edge {
    long long to, time, money;
    Edge(long long _to, long long _money, long long _time) {
        to = _to;
        time = _time;
        money = _money;
    }
};

int main() {
    int N;
    cin >> N;
    vector<long long> a(N);
    long long totalXor = 0;
    for (int i = 0; i < N; i++) {
        long long temp;
        cin >> temp;
        a[i] = temp;
        totalXor ^= temp;
    }

    for (int i = 0; i< N; i++) {
        cout << (a[i] ^ totalXor) << " ";
    }
    cout << endl;
    return 0;
}