#include <bits/stdc++.h>
const int N = 2e5, M = 64;
int n, cnt[M + 5];
long long a[N + 5], b[N + 5], c[N + 5];
int main() {
  scanf("%d", &n);
  long long sum = 0;
  for (int i = 1; i <= n; i++) scanf("%lld", &b[i]), sum += b[i];
  for (int i = 1; i <= n; i++) scanf("%lld", &c[i]), sum += c[i];
  if (sum % (2 * n) != 0) {
    puts("-1");
    return 0;
  }
  sum /= (2 * n);
  for (int i = 1; i <= n; i++) {
    long long tmp = b[i] + c[i] - sum;
    if (tmp < 0 || tmp % n != 0) {
      puts("-1");
      return 0;
    }
    a[i] = tmp / n;
    for (int j = 0; j <= 63; j++)
      if (((a[i] >> j) & 1) == 1) cnt[j]++;
  }
  for (int i = 1; i <= n; i++) {
    long long tmp1 = 0, tmp2 = 0;
    for (int j = 0; j <= 63; j++) {
      if (((a[i] >> j) & 1) == 1) {
        tmp1 += (1LL << j) * cnt[j];
        tmp2 += (1LL << j) * n;
      } else {
        tmp2 += (1LL << j) * cnt[j];
      }
    }
    if (tmp1 != b[i] || tmp2 != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld%c", a[i], " \n"[i == n]);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
const int N = 2e5, M = 64;
int n, cnt[M + 5];
long long a[N + 5], b[N + 5], c[N + 5];
int main() {
  scanf("%d", &n);
  long long sum = 0;
  for (int i = 1; i <= n; i++) scanf("%lld", &b[i]), sum += b[i];
  for (int i = 1; i <= n; i++) scanf("%lld", &c[i]), sum += c[i];
  if (sum % (2 * n) != 0) {
    puts("-1");
    return 0;
  }
  sum /= (2 * n);
  for (int i = 1; i <= n; i++) {
    long long tmp = b[i] + c[i] - sum;
    if (tmp < 0 || tmp % n != 0) {
      puts("-1");
      return 0;
    }
    a[i] = tmp / n;
    for (int j = 0; j <= 63; j++)
      if (((a[i] >> j) & 1) == 1) cnt[j]++;
  }
  for (int i = 1; i <= n; i++) {
    long long tmp1 = 0, tmp2 = 0;
    for (int j = 0; j <= 63; j++) {
      if (((a[i] >> j) & 1) == 1) {
        tmp1 += (1LL << j) * cnt[j];
        tmp2 += (1LL << j) * n;
      } else {
        tmp2 += (1LL << j) * cnt[j];
      }
    }
    if (tmp1 != b[i] || tmp2 != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld%c", a[i], " \n"[i == n]);
  return 0;
}

```