#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  long long b[n], c[n];
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  long long d[n], a[n], sum = 0;
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    sum += d[i];
  }
  sum /= (2 * n);
  bool flag = true;
  long long k[64];
  for (int j = 0; j < 63; j++) k[j] = 0;
  for (int i = 0; i < n; i++) {
    a[i] = (d[i] - sum) / n;
    if (a[i] < 0) {
      flag = false;
      break;
    }
    for (int j = 0; j < 63; j++)
      if (a[i] & (1LL << j)) k[j]++;
  }
  for (int i = 0; i < n && flag; i++) {
    long long num1 = 0, num2 = 0;
    for (int j = 0; j < 63; j++) {
      long long tmp = 1LL << j;
      if (a[i] & tmp) {
        num1 += k[j] << j;
        num2 += n << j;
      } else {
        num2 += k[j] << j;
      }
    }
    if (num1 != b[i] || num2 != c[i]) {
      flag = false;
      break;
    }
  }
  if (flag) {
    for (int i = 0; i < n; i++) cout << a[i] << " ";
    cout << endl;
  } else
    cout << "-1\n";
  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 main() {
  int n;
  cin >> n;
  long long b[n], c[n];
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  long long d[n], a[n], sum = 0;
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    sum += d[i];
  }
  sum /= (2 * n);
  bool flag = true;
  long long k[64];
  for (int j = 0; j < 63; j++) k[j] = 0;
  for (int i = 0; i < n; i++) {
    a[i] = (d[i] - sum) / n;
    if (a[i] < 0) {
      flag = false;
      break;
    }
    for (int j = 0; j < 63; j++)
      if (a[i] & (1LL << j)) k[j]++;
  }
  for (int i = 0; i < n && flag; i++) {
    long long num1 = 0, num2 = 0;
    for (int j = 0; j < 63; j++) {
      long long tmp = 1LL << j;
      if (a[i] & tmp) {
        num1 += k[j] << j;
        num2 += n << j;
      } else {
        num2 += k[j] << j;
      }
    }
    if (num1 != b[i] || num2 != c[i]) {
      flag = false;
      break;
    }
  }
  if (flag) {
    for (int i = 0; i < n; i++) cout << a[i] << " ";
    cout << endl;
  } else
    cout << "-1\n";
  return 0;
}

```