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:

<image>

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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.stream.LongStream;

public class Main {

    public static void main(String[] args) throws IOException {
        new Main().solve();
    }

    StringTokenizer tokenizer = null;
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

    String nextToken() throws IOException {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    void solve() throws IOException {
        int n = nextInt();
        long[] b = new long[n];
        long[] c = new long[n];
        for (int i = 0; i < n; i++) {
            b[i] = nextInt();
        }
        for (int i = 0; i < n; i++) {
            c[i] = nextInt();
        }
        long[] d = new long[n];
        for (int i = 0; i < n; i++) {
            d[i] = b[i] + c[i];
        }
        long dSum = LongStream.of(d).sum();
        long aSum = dSum / (2 * n);
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = (d[i] - aSum) / n;
        }

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

        long[] bToCheck = new long[n];
        long[] cToCheck = new long[n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 60; j++) {
                if ((a[i] & degrees[j]) > 0) {
                    bToCheck[i] += bits[j] * degrees[j];
                    cToCheck[i] += n * degrees[j];
                } else {
                    cToCheck[i] += bits[j] * degrees[j];
                }
            }
        }
        boolean bMatch = Arrays.equals(bToCheck, b);
        boolean cMatch = Arrays.equals(cToCheck, c);
        if (!bMatch || !cMatch) {
            System.out.println(-1);
        } else {
            for (int i = 0; i < n; i++) {
                System.out.print(a[i] + " ");
            }
        }
    }

}