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 <float.h>
#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

// 内部定数
#define D_ARRAY_MAX		200000									// 最大配列数
#define D_DCNT_MAX		30										// 最大桁数

// 内部変数
static FILE *szpFpI;											// 入力
static int si1Array[D_ARRAY_MAX];								// 配列
static int siACnt;												// 配列数
static int si1Res[D_ARRAY_MAX];									// 結果
static char sc1Out[D_ARRAY_MAX * 10];							// 出力内容
static char *scpOut;											// 出力位置

// 内部変数 - テスト用
#ifdef D_TEST
	static int siRes;
	static FILE *szpFpA;
	static int siTNo;
#endif

// １行出力
int
fOutLine(
	char *pcpLine				// <I> １行
)
{
	char lc1Buf[1024];

#ifdef D_TEST
	lc1Buf[0] = '\0';
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, pcpLine)) {
		siRes = -1;
	}
#else
	printf("%s", pcpLine);
#endif

	return 0;
}

// 出力内容 - 追加
int
fOutAdd(
	int piVal					// <I> 値
)
{
	char lc1Buf[1024];
	sprintf(lc1Buf, "%d ", piVal);
	int liLen = strlen(lc1Buf);
	memcpy(scpOut, lc1Buf, liLen);
	scpOut += liLen;

	return 0;
}

// 実行メイン
int
fMain(
)
{
	int i, j;
	char lc1Buf[1024];

	// データ - 初期化
	memset(si1Res, 0, sizeof(si1Res));
	scpOut = sc1Out;

	// 配列数 - 取得
	fgets(lc1Buf, sizeof(lc1Buf), szpFpI);
	sscanf(lc1Buf, "%d", &siACnt);

	// 配列 - 取得
	for (i = 0; i < siACnt; i++) {
		fscanf(szpFpI, "%d", &si1Array[i]);
	}
	fgets(lc1Buf, sizeof(lc1Buf), szpFpI);

	// 桁数別にセット
	for (i = 0; i < D_DCNT_MAX; i++) {
		int liTgt = 1 << i;

		// 個数 - 取得
		int liCnt = 0;
		for (j = 0; j < siACnt; j++) {
			if (liTgt & si1Array[j]) {
				liCnt++;
			}
		}
		if (liCnt == 0) {
			continue;
		}

		// 結果 - 加算
		liCnt = liTgt * ((liCnt + 1) % 2);
		for (j = 0; j < siACnt; j++) {
			if ((liTgt & si1Array[j]) == liCnt) {
				si1Res[j] += liTgt;
			}
		}
	}

	// 出力
	for (i = 0; i < siACnt; i++) {
		fOutAdd(si1Res[i]);
	}
	*scpOut = '\0';
	scpOut--;
	*scpOut = '\n';
	fOutLine(sc1Out);

	return 0;
}

// １回実行
int
fOne(
)
{
	int liRet;
	char lc1Buf[1024];

	// 入力 - セット
#ifdef D_TEST
	sprintf(lc1Buf, ".\\Test\\T%d.txt", siTNo);
	szpFpI = fopen(lc1Buf, "r");
	sprintf(lc1Buf, ".\\Test\\A%d.txt", siTNo);
	szpFpA = fopen(lc1Buf, "r");
	siRes = 0;
#else
	szpFpI = stdin;
#endif

	// 実行メイン
	liRet = fMain();

	// 残データ有無
#ifdef D_TEST
	lc1Buf[0] = '\0';
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, "")) {
		siRes = -1;
	}
#endif

	// テストファイルクローズ
#ifdef D_TEST
	fclose(szpFpI);
	fclose(szpFpA);
#endif

	// テスト結果
#ifdef D_TEST
	if (siRes == 0) {
		printf("OK %d\n", siTNo);
	}
	else {
		printf("NG %d\n", siTNo);
	}
#endif

	return 0;
}

// プログラム開始
int
main()
{

#ifdef D_TEST
	int i;
	for (i = D_TEST_SNO; i <= D_TEST_ENO; i++) {
		siTNo = i;
		fOne();
	}
#else
	fOne();
#endif

	return 0;
}