#include <bits/stdc++.h>
using namespace std;
int n;
long long ans[200005], b[200005], c[200005], sum, cnt[70], tmp;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%lld", &b[i]);
    sum += b[i];
  }
  for (register int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
    sum += c[i];
  }
  if (sum % (n * 2) != 0) {
    printf("-1\n");
    return 0;
  }
  sum /= (n * 2);
  for (register int i = 1; i <= n; i++) {
    if ((b[i] + c[i]) % n != sum % n) {
      printf("-1\n");
      return 0;
    }
    ans[i] = (b[i] + c[i] - sum) / n;
    for (register int j = 1; j <= 63; j++) {
      cnt[j] += (ans[i] & (1ll << (j - 1))) > 0;
    }
  }
  for (register int i = 1; i <= n; i++) {
    tmp = 0;
    for (register int j = 1; j <= 63; j++) {
      tmp += cnt[j] * ((ans[i] & (1ll << (j - 1))) > 0) * (1ll << (j - 1));
    }
    if (b[i] != tmp) {
      printf("-1\n");
      return 0;
    }
    tmp = 0;
    for (register int j = 1; j <= 63; j++) {
      tmp +=
          (n - cnt[j]) * ((ans[i] & (1ll << (j - 1))) > 0) * (1ll << (j - 1));
    }
    if (c[i] != tmp + sum) {
      printf("-1\n");
      return 0;
    }
  }
  for (register int i = 1; i < n; i++) {
    printf("%lld ", ans[i]);
  }
  printf("%lld\n", ans[n]);
  return 0;
}

### Prompt

Generate a cpp solution 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;
long long ans[200005], b[200005], c[200005], sum, cnt[70], tmp;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%lld", &b[i]);
    sum += b[i];
  }
  for (register int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
    sum += c[i];
  }
  if (sum % (n * 2) != 0) {
    printf("-1\n");
    return 0;
  }
  sum /= (n * 2);
  for (register int i = 1; i <= n; i++) {
    if ((b[i] + c[i]) % n != sum % n) {
      printf("-1\n");
      return 0;
    }
    ans[i] = (b[i] + c[i] - sum) / n;
    for (register int j = 1; j <= 63; j++) {
      cnt[j] += (ans[i] & (1ll << (j - 1))) > 0;
    }
  }
  for (register int i = 1; i <= n; i++) {
    tmp = 0;
    for (register int j = 1; j <= 63; j++) {
      tmp += cnt[j] * ((ans[i] & (1ll << (j - 1))) > 0) * (1ll << (j - 1));
    }
    if (b[i] != tmp) {
      printf("-1\n");
      return 0;
    }
    tmp = 0;
    for (register int j = 1; j <= 63; j++) {
      tmp +=
          (n - cnt[j]) * ((ans[i] & (1ll << (j - 1))) > 0) * (1ll << (j - 1));
    }
    if (c[i] != tmp + sum) {
      printf("-1\n");
      return 0;
    }
  }
  for (register int i = 1; i < n; i++) {
    printf("%lld ", ans[i]);
  }
  printf("%lld\n", ans[n]);
  return 0;
}

```