#include <bits/stdc++.h>
using namespace std;
int n;
long long a[200200];
long long b[200200];
long long c[200200];
long long bits[40];
long long sum;
int main() {
  ios::sync_with_stdio(false);
  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 += b[i] + c[i];
  }
  sum /= 2;
  sum /= n;
  for (int i = 0; i < n; i++) {
    a[i] = ((b[i] + c[i]) - sum) / n;
  }
  for (int i = 0; i < 32; i++) {
    for (int j = 0; j < n; j++) {
      if (a[j] & (1 << i)) {
        bits[i]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    long long andy = 0, ore = 0;
    for (int j = 0; j < 32; j++) {
      if (a[i] & (1 << j)) {
        andy += (1 << j) * bits[j];
        ore += (1 << j) * n;
      } else {
        ore += (1 << j) * bits[j];
      }
    }
    if (andy != b[i] || ore != c[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    if (i) cout << " ";
    cout << a[i];
  }
  cout << 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 n;
long long a[200200];
long long b[200200];
long long c[200200];
long long bits[40];
long long sum;
int main() {
  ios::sync_with_stdio(false);
  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 += b[i] + c[i];
  }
  sum /= 2;
  sum /= n;
  for (int i = 0; i < n; i++) {
    a[i] = ((b[i] + c[i]) - sum) / n;
  }
  for (int i = 0; i < 32; i++) {
    for (int j = 0; j < n; j++) {
      if (a[j] & (1 << i)) {
        bits[i]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    long long andy = 0, ore = 0;
    for (int j = 0; j < 32; j++) {
      if (a[i] & (1 << j)) {
        andy += (1 << j) * bits[j];
        ore += (1 << j) * n;
      } else {
        ore += (1 << j) * bits[j];
      }
    }
    if (andy != b[i] || ore != c[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    if (i) cout << " ";
    cout << a[i];
  }
  cout << endl;
  return 0;
}

```