#include <bits/stdc++.h>
const int N = 200005;
long long s;
int n, b[N], c[N], a[N], cnt[40];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]), s += b[i];
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]), s += c[i];
  s = s / 2 / n;
  for (int i = 1; i <= n; i++) a[i] = (c[i] + b[i] - s) / n;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 31; j++)
      if (a[i] & (1 << j)) cnt[j]++;
  for (int i = 1; i <= n; i++) {
    int sum = 0;
    for (int j = 0; j < 31; j++)
      if (a[i] & (1 << j)) sum = sum + 1ll * cnt[j] * (1 << j);
    if (sum != b[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
}

### 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>
const int N = 200005;
long long s;
int n, b[N], c[N], a[N], cnt[40];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]), s += b[i];
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]), s += c[i];
  s = s / 2 / n;
  for (int i = 1; i <= n; i++) a[i] = (c[i] + b[i] - s) / n;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 31; j++)
      if (a[i] & (1 << j)) cnt[j]++;
  for (int i = 1; i <= n; i++) {
    int sum = 0;
    for (int j = 0; j < 31; j++)
      if (a[i] & (1 << j)) sum = sum + 1ll * cnt[j] * (1 << j);
    if (sum != b[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
}

```