#include <bits/stdc++.h>
using namespace std;
int n;
int b[200005], c[200005];
long long tot;
int a[200005];
int sum[32];
bool check() {
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 30; j++)
      if (a[i] & (1 << j)) sum[j]++;
  }
  for (int i = 1; i <= n; i++) {
    long long nowb = 0, nowc = 0;
    for (int j = 0; j <= 30; j++) {
      if (a[i] & (1 << j)) {
        nowb += sum[j] * (1 << j);
        nowc += n * (1 << j);
      } else {
        nowc += sum[j] * (1 << j);
      }
    }
    if (nowb != b[i] || nowc != c[i]) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]), tot += b[i];
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]), tot += c[i];
  if (tot % (2 * n) != 0) {
    puts("-1");
    return 0;
  }
  tot /= (2 * n);
  for (int i = 1; i <= n; i++) {
    long long tmp = b[i] + c[i] - tot;
    if (tmp < 0 || tmp % n != 0) {
      puts("-1");
      return 0;
    }
    a[i] = tmp / n;
  }
  if (!check()) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int b[200005], c[200005];
long long tot;
int a[200005];
int sum[32];
bool check() {
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 30; j++)
      if (a[i] & (1 << j)) sum[j]++;
  }
  for (int i = 1; i <= n; i++) {
    long long nowb = 0, nowc = 0;
    for (int j = 0; j <= 30; j++) {
      if (a[i] & (1 << j)) {
        nowb += sum[j] * (1 << j);
        nowc += n * (1 << j);
      } else {
        nowc += sum[j] * (1 << j);
      }
    }
    if (nowb != b[i] || nowc != c[i]) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]), tot += b[i];
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]), tot += c[i];
  if (tot % (2 * n) != 0) {
    puts("-1");
    return 0;
  }
  tot /= (2 * n);
  for (int i = 1; i <= n; i++) {
    long long tmp = b[i] + c[i] - tot;
    if (tmp < 0 || tmp % n != 0) {
      puts("-1");
      return 0;
    }
    a[i] = tmp / n;
  }
  if (!check()) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}

```