#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100, LOG = 30;
void bad() {
  cout << -1;
  exit(0);
}
int n, a[N], b[N], c[N], cnt[LOG];
long long s;
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    s += b[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    s += c[i];
  }
  s /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = (b[i] + c[i] - s) / n;
    if (a[i] < 0) bad();
    for (int j = 0; j < LOG; j++) cnt[j] += (a[i] >> j) & 1;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < LOG; j++)
      if (a[i] & (1 << j)) {
        b[i] -= (cnt[j] << j);
        c[i] -= n << j;
      } else
        c[i] -= cnt[j] << j;
    if (b[i] || c[i]) bad();
  }
  for (int i = 0; i < n; i++) cout << a[i] << ' ';
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const int N = 2e5 + 100, LOG = 30;
void bad() {
  cout << -1;
  exit(0);
}
int n, a[N], b[N], c[N], cnt[LOG];
long long s;
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    s += b[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    s += c[i];
  }
  s /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = (b[i] + c[i] - s) / n;
    if (a[i] < 0) bad();
    for (int j = 0; j < LOG; j++) cnt[j] += (a[i] >> j) & 1;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < LOG; j++)
      if (a[i] & (1 << j)) {
        b[i] -= (cnt[j] << j);
        c[i] -= n << j;
      } else
        c[i] -= cnt[j] << j;
    if (b[i] || c[i]) bad();
  }
  for (int i = 0; i < n; i++) cout << a[i] << ' ';
  return 0;
}

```