#include <bits/stdc++.h>
using namespace std;
long long int B[210001], C[210001];
long long int A[210001], total[40];
int main() {
  long long int N;
  cin >> N;
  long long int sum = 0;
  for (int i = 0; i < N; i++) cin >> B[i], sum += B[i];
  for (int i = 0; i < N; i++) cin >> C[i], sum += C[i];
  if (sum % 2 * N != 0) {
    cout << "-1\n";
    return 0;
  }
  long long int d = sum / (2 * N);
  for (int i = 0; i < N; i++) {
    long long int cur = B[i] + C[i] - d;
    if (cur % N != 0) {
      cout << "-1\n";
      return 0;
    }
    A[i] = (cur) / N;
    for (int k = 0; (1LL << k) <= A[i]; k++) {
      if (A[i] & (1LL << k)) total[k] += (1LL << k);
    }
  }
  for (int i = 0; i < N; i++) {
    long long int and_sum = 0;
    long long int or_sum = 0;
    for (long long int k = 0; k < 40; k++) {
      if (A[i] & (1LL << k)) {
        and_sum += total[k];
        or_sum += (1LL << k) * N;
      } else {
        or_sum += total[k];
      }
    }
    if (or_sum != C[i] || and_sum != B[i]) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 0; i < N; i++) cout << A[i] << " ";
  cout << "\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;
long long int B[210001], C[210001];
long long int A[210001], total[40];
int main() {
  long long int N;
  cin >> N;
  long long int sum = 0;
  for (int i = 0; i < N; i++) cin >> B[i], sum += B[i];
  for (int i = 0; i < N; i++) cin >> C[i], sum += C[i];
  if (sum % 2 * N != 0) {
    cout << "-1\n";
    return 0;
  }
  long long int d = sum / (2 * N);
  for (int i = 0; i < N; i++) {
    long long int cur = B[i] + C[i] - d;
    if (cur % N != 0) {
      cout << "-1\n";
      return 0;
    }
    A[i] = (cur) / N;
    for (int k = 0; (1LL << k) <= A[i]; k++) {
      if (A[i] & (1LL << k)) total[k] += (1LL << k);
    }
  }
  for (int i = 0; i < N; i++) {
    long long int and_sum = 0;
    long long int or_sum = 0;
    for (long long int k = 0; k < 40; k++) {
      if (A[i] & (1LL << k)) {
        and_sum += total[k];
        or_sum += (1LL << k) * N;
      } else {
        or_sum += total[k];
      }
    }
    if (or_sum != C[i] || and_sum != B[i]) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 0; i < N; i++) cout << A[i] << " ";
  cout << "\n";
  return 0;
}

```