#include <bits/stdc++.h>
using namespace std;
int N, b[200009], c[200009], cnt[50];
long long S, a[200009];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &b[i]);
  for (int i = 1; i <= N; i++) scanf("%d", &c[i]), S += b[i] + c[i];
  if (S % (2 * N) != 0) {
    printf("-1\n");
    return 0;
  }
  S /= (2 * N);
  for (int i = 1; i <= N; i++) a[i] = (b[i] + c[i] - S) / N;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j <= 40; j++) cnt[j] += ((a[i] & (1LL << j)) > 0);
  for (int i = 1; i <= N; i++) {
    long long an = 0, sa = 0;
    for (int j = 0; j <= 40; j++) {
      if (a[i] & (1LL << j))
        an += (1LL << j) * cnt[j], sa += (1LL << j) * N;
      else
        sa += (1LL << j) * cnt[j];
    }
    if (an != b[i] || sa != c[i]) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i <= N; i++) printf("%I64d ", a[i]);
  printf("\n");
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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>
using namespace std;
int N, b[200009], c[200009], cnt[50];
long long S, a[200009];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &b[i]);
  for (int i = 1; i <= N; i++) scanf("%d", &c[i]), S += b[i] + c[i];
  if (S % (2 * N) != 0) {
    printf("-1\n");
    return 0;
  }
  S /= (2 * N);
  for (int i = 1; i <= N; i++) a[i] = (b[i] + c[i] - S) / N;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j <= 40; j++) cnt[j] += ((a[i] & (1LL << j)) > 0);
  for (int i = 1; i <= N; i++) {
    long long an = 0, sa = 0;
    for (int j = 0; j <= 40; j++) {
      if (a[i] & (1LL << j))
        an += (1LL << j) * cnt[j], sa += (1LL << j) * N;
      else
        sa += (1LL << j) * cnt[j];
    }
    if (an != b[i] || sa != c[i]) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i <= N; i++) printf("%I64d ", a[i]);
  printf("\n");
  return 0;
}

```