#include <bits/stdc++.h>
using namespace std;
long long n, sum, a[200005], b[200005], c[200005], cnt[30];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) cin >> c[i];
  for (long long i = 1; i <= n; i++) sum += b[i] + c[i];
  if (sum % (n * 2)) {
    cout << -1;
    return 0;
  }
  sum /= n * 2;
  for (long long i = 1; i <= n; i++) {
    if ((b[i] + c[i] - sum) % n) {
      cout << -1;
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  for (long long j = 0; j <= 29; j++) {
    for (long long i = 1; i <= n; i++) cnt[j] += ((a[i] & (1 << j)) > 0);
  }
  for (long long i = 1; i <= n; i++) {
    long long r = 0;
    for (long long j = 0; j <= 29; j++) {
      if (a[i] & (1 << j))
        r += n * (1 << j);
      else
        r += cnt[j] * (1 << j);
    }
    if (c[i] != r) {
      cout << -1;
      return 0;
    }
  }
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
}

### Prompt

Your task is to create 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 n, sum, a[200005], b[200005], c[200005], cnt[30];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) cin >> c[i];
  for (long long i = 1; i <= n; i++) sum += b[i] + c[i];
  if (sum % (n * 2)) {
    cout << -1;
    return 0;
  }
  sum /= n * 2;
  for (long long i = 1; i <= n; i++) {
    if ((b[i] + c[i] - sum) % n) {
      cout << -1;
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  for (long long j = 0; j <= 29; j++) {
    for (long long i = 1; i <= n; i++) cnt[j] += ((a[i] & (1 << j)) > 0);
  }
  for (long long i = 1; i <= n; i++) {
    long long r = 0;
    for (long long j = 0; j <= 29; j++) {
      if (a[i] & (1 << j))
        r += n * (1 << j);
      else
        r += cnt[j] * (1 << j);
    }
    if (c[i] != r) {
      cout << -1;
      return 0;
    }
  }
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
}

```