#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
unsigned long long Total;
unsigned long long Array[200001];
unsigned long long Brray[200001];
unsigned long long Crray[200001];
unsigned long long Count[101];
int main(void) {
  register unsigned long long i, j;
  cin >> Total;
  for (i = 1; i <= Total; i++) {
    cin >> Brray[i];
  }
  for (i = 1; i <= Total; i++) {
    cin >> Crray[i];
  }
  register unsigned long long Sum;
  Sum = 0;
  for (i = 1; i <= Total; i++) {
    Sum += Brray[i];
  }
  for (i = 1; i <= Total; i++) {
    Sum += Crray[i];
  }
  if (Sum % (Total * 2) != 0) {
    goto Impossible;
  }
  Sum /= (Total * 2);
  for (i = 1; i <= Total; i++) {
    Array[i] = Brray[i] + Crray[i] - Sum;
    if (Array[i] % Total != 0) {
      goto Impossible;
    }
    Array[i] /= Total;
    for (j = 1; j <= 63; j++) {
      if ((Array[i] & (unsigned long long)pow(2, j - 1)) != 0) {
        Count[j]++;
      }
    }
  }
  for (i = 1; i <= Total; i++) {
    register unsigned long long Temp;
    Temp = 0;
    for (j = 1; j <= 63; j++) {
      if ((Array[i] & (unsigned long long)pow(2, j - 1)) != 0) {
        Temp += Count[j] * (unsigned long long)pow(2, j - 1);
      }
    }
    if (Temp != Brray[i]) {
      goto Impossible;
    }
    Temp = Sum;
    for (j = 1; j <= 63; j++) {
      if ((Array[i] & (unsigned long long)pow(2, j - 1)) != 0) {
        Temp += (Total - Count[j]) * (unsigned long long)pow(2, j - 1);
      }
    }
    if (Temp != Crray[i]) {
      goto Impossible;
    }
  }
  for (i = 1; i <= Total; i++) {
    cout << Array[i] << ' ';
  }
  cout << endl;
  return 0;
Impossible:
  cout << -1 << endl;
  return 0;
}

### 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>
#pragma GCC optimize(3)
using namespace std;
unsigned long long Total;
unsigned long long Array[200001];
unsigned long long Brray[200001];
unsigned long long Crray[200001];
unsigned long long Count[101];
int main(void) {
  register unsigned long long i, j;
  cin >> Total;
  for (i = 1; i <= Total; i++) {
    cin >> Brray[i];
  }
  for (i = 1; i <= Total; i++) {
    cin >> Crray[i];
  }
  register unsigned long long Sum;
  Sum = 0;
  for (i = 1; i <= Total; i++) {
    Sum += Brray[i];
  }
  for (i = 1; i <= Total; i++) {
    Sum += Crray[i];
  }
  if (Sum % (Total * 2) != 0) {
    goto Impossible;
  }
  Sum /= (Total * 2);
  for (i = 1; i <= Total; i++) {
    Array[i] = Brray[i] + Crray[i] - Sum;
    if (Array[i] % Total != 0) {
      goto Impossible;
    }
    Array[i] /= Total;
    for (j = 1; j <= 63; j++) {
      if ((Array[i] & (unsigned long long)pow(2, j - 1)) != 0) {
        Count[j]++;
      }
    }
  }
  for (i = 1; i <= Total; i++) {
    register unsigned long long Temp;
    Temp = 0;
    for (j = 1; j <= 63; j++) {
      if ((Array[i] & (unsigned long long)pow(2, j - 1)) != 0) {
        Temp += Count[j] * (unsigned long long)pow(2, j - 1);
      }
    }
    if (Temp != Brray[i]) {
      goto Impossible;
    }
    Temp = Sum;
    for (j = 1; j <= 63; j++) {
      if ((Array[i] & (unsigned long long)pow(2, j - 1)) != 0) {
        Temp += (Total - Count[j]) * (unsigned long long)pow(2, j - 1);
      }
    }
    if (Temp != Crray[i]) {
      goto Impossible;
    }
  }
  for (i = 1; i <= Total; i++) {
    cout << Array[i] << ' ';
  }
  cout << endl;
  return 0;
Impossible:
  cout << -1 << endl;
  return 0;
}

```