734_F. Anton and School_13657
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
#include <bits/stdc++.h>
int n;
long long a[(200010)], b[(200010)], c[(200010)], cnt[(31)];
int main() {
  srand((unsigned)time(NULL));
  scanf("%d", &n);
  long long sum = 0, suma;
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", b + i);
    sum += b[i];
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", c + i);
    sum += c[i];
  }
  if (sum % (2 * n)) return printf("-1\n"), 0;
  suma = sum / (2 * n);
  for (int i = 1; i <= n; ++i) {
    if ((b[i] + c[i] - suma) % n) return printf("-1\n"), 0;
    a[i] = (b[i] + c[i] - suma) / n;
    for (int j = 0; j < (31); ++j) {
      cnt[j] += (a[i] >> j) & 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    long long sumb = 0, sumc = 0;
    for (int j = 0; j < (31); ++j) {
      if ((a[i] >> j) & 1) {
        sumb += (1ll << j) * cnt[j];
        sumc += (1ll << j) * n;
      } else {
        sumc += (1ll << j) * cnt[j];
      }
    }
    if (sumb != b[i] || sumc != c[i]) return printf("-1\n"), 0;
  }
  for (int i = 1; i <= n; ++i) {
    printf("%I64d%c", a[i], " \n"[i == n]);
  }
  return 0;
}
