#include <bits/stdc++.h>
using namespace std;
long long b[200500], c[200500];
long long n;
long long a[200500];
long long sa;
long long s[33];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    sa += b[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    sa += c[i];
  }
  if (sa % (n + n) != 0) {
    cout << -1;
    return 0;
  }
  sa = sa / (n + n);
  for (int i = 0; i < n; i++) {
    a[i] = b[i] + c[i] - sa;
    if (a[i] % n != 0) {
      cout << -1;
      return 0;
    }
    a[i] = a[i] / n;
    if (a[i] < 0) {
      cout << -1;
      return 0;
    }
    for (int j = 0; j < 31; j++) {
      if ((a[i] & (1 << j)) != 0) {
        s[j]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    long long a1 = 0;
    for (int j = 0; j < 31; j++) {
      if ((a[i] & (1 << j)) != 0) {
        a1 = a1 + s[j] * (long long)(1 << j);
      }
    }
    long long a2 = 0;
    for (int j = 0; j < 31; j++) {
      if ((a[i] & (1 << j)) == 0) {
        a2 = a2 + s[j] * (long long)(1 << j);
      } else {
        a2 = a2 + n * (long long)(1 << j);
      }
    }
    if (b[i] != a1 || c[i] != a2) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long b[200500], c[200500];
long long n;
long long a[200500];
long long sa;
long long s[33];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    sa += b[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    sa += c[i];
  }
  if (sa % (n + n) != 0) {
    cout << -1;
    return 0;
  }
  sa = sa / (n + n);
  for (int i = 0; i < n; i++) {
    a[i] = b[i] + c[i] - sa;
    if (a[i] % n != 0) {
      cout << -1;
      return 0;
    }
    a[i] = a[i] / n;
    if (a[i] < 0) {
      cout << -1;
      return 0;
    }
    for (int j = 0; j < 31; j++) {
      if ((a[i] & (1 << j)) != 0) {
        s[j]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    long long a1 = 0;
    for (int j = 0; j < 31; j++) {
      if ((a[i] & (1 << j)) != 0) {
        a1 = a1 + s[j] * (long long)(1 << j);
      }
    }
    long long a2 = 0;
    for (int j = 0; j < 31; j++) {
      if ((a[i] & (1 << j)) == 0) {
        a2 = a2 + s[j] * (long long)(1 << j);
      } else {
        a2 = a2 + n * (long long)(1 << j);
      }
    }
    if (b[i] != a1 || c[i] != a2) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
}

```