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.math.BigDecimal;
import java.util.*;

/**
 * Created by chao on 16/11/13.
 */
public class A {


    public static void main(String[] args) throws IOException {
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int n = Integer.parseInt(br.readLine());
        String[] nums = br.readLine().split(" ");
        long[] b = new long[n];
        long sb = 0;
        for (int i = 0; i < n; i++) {
            b[i] = Long.parseLong(nums[i]);
            sb += b[i];
        }
        nums = br.readLine().split(" ");
        long[] c = new long[n];
        long sc = 0;
        for (int i = 0; i < n; i++) {
            c[i] = Long.parseLong(nums[i]);
            sc += c[i];
        }

        long[] a = new long[n];
        long sa = 0;
        for (int i = 0; i < n; i++) {
            a[i] = (b[i] + c[i] - (sb + sc)/2/n)/n;
            sa += a[i];
        }

        int[] bits = new int[64];
        for (int i = 0; i < n; i++) {
            for (int k = 0; k < 60; k++) {
                if (((1L<<k)&a[i]) > 0) {
                    bits[k]++;
                }
            }
        }


        boolean isOK = true;
        for (int i = 0; i < n; i++) {
            if (b[i] != n*a[i] +sa - c[i]) {
                isOK = false;
                break;
            }
        }
        if (isOK) {
            for (int i = 0; i < n; i++) {
                if (b[i] != cal(bits, a[i])) {
                    isOK = false;
                    break;
                }
            }
        }

        if (!isOK) pw.println(-1);
        else {
            for (int i = 0; i < n; i++) {
                if (i != 0) pw.print(' ');
                pw.print(a[i]);
            }
            pw.println();
        }



        pw.flush();
        pw.close();
        br.close();

    }

    private static long cal(int[] bits, long a) {
        long ans = 0;
        for (int i = 0; i < 60; i++) {
            if (((1L<<i)&a) > 0) ans += bits[i]*(1L<<i);
        }
        return ans;
    }
}

