Anton goes to school, his favorite lessons are arraystudying. He usually solves all the tasks pretty fast, but this time the teacher gave him a complicated one: given two arrays b and c of length n, find array a, such that:

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where a and b means bitwise AND, while a or b means bitwise OR.

Usually Anton is good in arraystudying, but this problem is too hard, so Anton asks you to help.

Input

The first line of the input contains a single integers n (1 ≤ n ≤ 200 000) — the size of arrays b and c.

The second line contains n integers bi (0 ≤ bi ≤ 109) — elements of the array b.

Third line contains n integers ci (0 ≤ ci ≤ 109) — elements of the array c.

Output

If there is no solution, print  - 1.

Otherwise, the only line of the output should contain n non-negative integers ai — elements of the array a. If there are multiple possible solutions, you may print any of them.

Examples

Input

4
6 8 4 4
16 22 10 10


Output

3 5 1 1 


Input

5
8 25 14 7 16
19 6 9 4 25


Output

-1
import java.io.*;
import java.util.*;

public class CF734F {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		int[] bb = new int[n];
		for (int i = 0; i < n; i++)
			bb[i] = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(br.readLine());
		int[] cc = new int[n];
		for (int i = 0; i < n; i++)
			cc[i] = Integer.parseInt(st.nextToken());
		long sum = 0;
		for (int i = 0; i < n; i++)
			sum += bb[i] + cc[i];
		if (sum % (2 * n) != 0) {
			System.out.println(-1);
			return;
		}
		sum /= 2 * n;
		int[] aa = new int[n];
		for (int i = 0; i < n; i++) {
			int bc = bb[i] + cc[i];
			if (bc < sum) {
				System.out.println(-1);
				return;
			}
			int d = (int) (bc - sum);
			if (d % n != 0) {
				System.out.println(-1);
				return;
			}
			aa[i] = d / n;
		}
		int[] kk = new int[30];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < 30; j++)
				if ((aa[i] & 1 << j) > 0)
					kk[j]++;
		for (int i = 0; i < n; i++) {
			long b = 0, c = 0;
			for (int j = 0; j < 30; j++) {
				long p = 1L << j;
				if ((aa[i] & 1 << j) > 0) {
					b += p * kk[j];
					c += p * n;
				} else
					c += p * kk[j];
			}
			if (bb[i] != b || cc[i] != c) {
				System.out.println(-1);
				return;
			}
		}
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < n; i++)
			sb.append(aa[i] + " ");
		System.out.println(sb);
	}
}
