#include <bits/stdc++.h>
long long a[212121], b[212121], c[212121], bb[212121], cc[212121], cnt[32];
int main() {
  long long n, i, j;
  scanf("%lld", &n);
  long long sum = 0, all;
  for (i = 0; i < n; i++) scanf("%lld", &b[i]), sum += b[i];
  for (i = 0; i < n; i++) scanf("%lld", &c[i]), sum += c[i];
  if (sum % (2 * n) != 0) {
    printf("-1");
    return 0;
  }
  all = sum / (2 * n);
  for (i = 0; i < n; i++) {
    long long ss = b[i] + c[i] - all;
    if (ss < 0 || ss % n != 0) {
      printf("-1");
      return 0;
    }
    a[i] = ss / n;
  }
  int ck = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < 31; j++) {
      if (a[i] & (1 << j)) cnt[j]++;
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < 31; j++) {
      if (a[i] & (1 << j)) bb[i] += cnt[j] * (1 << j);
    }
    cc[i] = n * a[i] + all - b[i];
    if (bb[i] != b[i] || cc[i] != c[i]) {
      printf("-1");
      return 0;
    }
  }
  for (i = 0; i < n; i++) printf("%lld ", a[i]);
  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>
long long a[212121], b[212121], c[212121], bb[212121], cc[212121], cnt[32];
int main() {
  long long n, i, j;
  scanf("%lld", &n);
  long long sum = 0, all;
  for (i = 0; i < n; i++) scanf("%lld", &b[i]), sum += b[i];
  for (i = 0; i < n; i++) scanf("%lld", &c[i]), sum += c[i];
  if (sum % (2 * n) != 0) {
    printf("-1");
    return 0;
  }
  all = sum / (2 * n);
  for (i = 0; i < n; i++) {
    long long ss = b[i] + c[i] - all;
    if (ss < 0 || ss % n != 0) {
      printf("-1");
      return 0;
    }
    a[i] = ss / n;
  }
  int ck = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < 31; j++) {
      if (a[i] & (1 << j)) cnt[j]++;
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < 31; j++) {
      if (a[i] & (1 << j)) bb[i] += cnt[j] * (1 << j);
    }
    cc[i] = n * a[i] + all - b[i];
    if (bb[i] != b[i] || cc[i] != c[i]) {
      printf("-1");
      return 0;
    }
  }
  for (i = 0; i < n; i++) printf("%lld ", a[i]);
  return 0;
}

```