#include <bits/stdc++.h>
using namespace std;
int sum[105], b[200050], c[200050], d[200050];
long long a[200050], sb[200050], sc[200050];
int main() {
  int n, i, j;
  long long sd, sa;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (i = 1; i <= n; i++) d[i] = b[i] + c[i];
  sd = sa = 0;
  for (i = 1; i <= n; i++) sd += d[i];
  sa = sd / n / 2;
  for (i = 1; i <= n; i++) a[i] = (d[i] - sa) / n;
  for (i = 0; i <= 30; i++)
    for (j = 1; j <= n; j++) sum[i] += (a[j] >> i) & 1;
  for (i = 0; i <= 30; i++) {
    for (j = 1; j <= n; j++)
      if ((1 << i) & a[j]) sb[j] += (1ll << i) * sum[i];
    for (j = 1; j <= n; j++)
      if ((1 << i) & a[j])
        sc[j] += (1ll << i) * n;
      else
        sc[j] += (1ll << i) * sum[i];
  }
  for (i = 1; i <= n; i++)
    if (b[i] != sb[i] || c[i] != sc[i]) {
      printf("-1\n");
      return 0;
    }
  for (i = 1; i <= n; i++) printf("%I64d ", a[i]);
  printf("\n");
  return 0;
}

### Prompt

Your task is to create 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 sum[105], b[200050], c[200050], d[200050];
long long a[200050], sb[200050], sc[200050];
int main() {
  int n, i, j;
  long long sd, sa;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (i = 1; i <= n; i++) d[i] = b[i] + c[i];
  sd = sa = 0;
  for (i = 1; i <= n; i++) sd += d[i];
  sa = sd / n / 2;
  for (i = 1; i <= n; i++) a[i] = (d[i] - sa) / n;
  for (i = 0; i <= 30; i++)
    for (j = 1; j <= n; j++) sum[i] += (a[j] >> i) & 1;
  for (i = 0; i <= 30; i++) {
    for (j = 1; j <= n; j++)
      if ((1 << i) & a[j]) sb[j] += (1ll << i) * sum[i];
    for (j = 1; j <= n; j++)
      if ((1 << i) & a[j])
        sc[j] += (1ll << i) * n;
      else
        sc[j] += (1ll << i) * sum[i];
  }
  for (i = 1; i <= n; i++)
    if (b[i] != sb[i] || c[i] != sc[i]) {
      printf("-1\n");
      return 0;
    }
  for (i = 1; i <= n; i++) printf("%I64d ", a[i]);
  printf("\n");
  return 0;
}

```