#include <bits/stdc++.h>
using namespace std;
int b[200010], c[200010], a[200010];
int bit[32];
int main() {
  ios::sync_with_stdio(false);
  int n;
  long long sum = 0;
  bool flag = true;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  for (int i = 0; i < n; i++) sum += (long long)b[i] + (long long)c[i];
  if (sum % (2 * n) != 0)
    flag = false;
  else {
    sum /= 2 * n;
    for (int i = 0; i < n; i++) {
      a[i] = (b[i] + c[i] - sum);
      if (a[i] % n != 0) {
        flag = false;
        break;
      } else
        a[i] /= n;
    }
    if (flag) {
      for (int i = 0; i < n; i++)
        for (int j = 0; j < 32; j++)
          if (a[i] & (1 << j)) bit[j]++;
      for (int i = 0, op; i < n; i++) {
        op = 0;
        for (int j = 0; j < 32; j++)
          if (a[i] & (1 << j)) op += bit[j] * (1 << j);
        if (op != b[i]) {
          flag = false;
          break;
        }
      }
    }
  }
  if (flag) {
    for (int i = 0; i < n; i++) cout << a[i] << " \n"[i == n - 1];
  } else
    cout << -1 << endl;
  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 b[200010], c[200010], a[200010];
int bit[32];
int main() {
  ios::sync_with_stdio(false);
  int n;
  long long sum = 0;
  bool flag = true;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  for (int i = 0; i < n; i++) sum += (long long)b[i] + (long long)c[i];
  if (sum % (2 * n) != 0)
    flag = false;
  else {
    sum /= 2 * n;
    for (int i = 0; i < n; i++) {
      a[i] = (b[i] + c[i] - sum);
      if (a[i] % n != 0) {
        flag = false;
        break;
      } else
        a[i] /= n;
    }
    if (flag) {
      for (int i = 0; i < n; i++)
        for (int j = 0; j < 32; j++)
          if (a[i] & (1 << j)) bit[j]++;
      for (int i = 0, op; i < n; i++) {
        op = 0;
        for (int j = 0; j < 32; j++)
          if (a[i] & (1 << j)) op += bit[j] * (1 << j);
        if (op != b[i]) {
          flag = false;
          break;
        }
      }
    }
  }
  if (flag) {
    for (int i = 0; i < n; i++) cout << a[i] << " \n"[i == n - 1];
  } else
    cout << -1 << endl;
  return 0;
}

```