#include <bits/stdc++.h>
using namespace std;
int a[200005], b[200005], ans[200005], n, tong[35];
int main() {
  cin >> n;
  bool flag = true;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  long long sum = 0;
  for (int i = 1; i <= n; i++) sum += a[i] + b[i];
  long long sigma_ans = sum / 2 / n;
  for (int i = 1; i <= n; i++) ans[i] = (a[i] + b[i] - sigma_ans) / n;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= 30; j++)
      if (ans[i] >> j & 1) tong[j]++;
  for (int i = 1; i <= n; i++) {
    long long now1 = 0, now2 = 0;
    for (int j = 0; j <= 30; j++) {
      if (ans[i] >> j & 1)
        now1 += 1ll * n * (1 << j), now2 += 1ll * tong[j] * (1 << j);
      else
        now1 += 1ll * tong[j] * (1 << j);
    }
    if (now2 != a[i] || now1 != b[i]) flag = false;
    if (!flag) break;
  }
  if (!flag) {
    printf("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Please formulate 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 a[200005], b[200005], ans[200005], n, tong[35];
int main() {
  cin >> n;
  bool flag = true;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  long long sum = 0;
  for (int i = 1; i <= n; i++) sum += a[i] + b[i];
  long long sigma_ans = sum / 2 / n;
  for (int i = 1; i <= n; i++) ans[i] = (a[i] + b[i] - sigma_ans) / n;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= 30; j++)
      if (ans[i] >> j & 1) tong[j]++;
  for (int i = 1; i <= n; i++) {
    long long now1 = 0, now2 = 0;
    for (int j = 0; j <= 30; j++) {
      if (ans[i] >> j & 1)
        now1 += 1ll * n * (1 << j), now2 += 1ll * tong[j] * (1 << j);
      else
        now1 += 1ll * tong[j] * (1 << j);
    }
    if (now2 != a[i] || now1 != b[i]) flag = false;
    if (!flag) break;
  }
  if (!flag) {
    printf("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}

```