#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
long long b[maxn], c[maxn], sum, num[maxn], a[maxn], s[maxn];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    sum += b[i] + c[i], num[i] = b[i] + c[i];
  }
  if (sum % (2 * n)) return cout << -1, 0;
  sum /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = num[i] - sum;
    if ((a[i]) % n || a[i] < 0) return cout << -1, 0;
    a[i] /= n;
  }
  for (int i = 0; i <= 30; i++)
    for (int j = 0; j < n; j++) s[i] += a[j] & (1 << i);
  long long k;
  for (int i = 0; i < n; i++) {
    k = 0;
    for (int j = 0; j <= 30; j++) {
      if (a[i] & (1 << j))
        k += n * (1 << j);
      else
        k += s[j];
    }
    if (k != c[i]) return cout << -1, 0;
    k = 0;
    for (int j = 0; j <= 30; j++)
      if (a[i] & (1 << j)) k += s[j];
    if (k != b[i]) return cout << -1, 0;
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
}

### Prompt

Please create a solution in cpp 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;
const int maxn = 2e5 + 10;
long long b[maxn], c[maxn], sum, num[maxn], a[maxn], s[maxn];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    sum += b[i] + c[i], num[i] = b[i] + c[i];
  }
  if (sum % (2 * n)) return cout << -1, 0;
  sum /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = num[i] - sum;
    if ((a[i]) % n || a[i] < 0) return cout << -1, 0;
    a[i] /= n;
  }
  for (int i = 0; i <= 30; i++)
    for (int j = 0; j < n; j++) s[i] += a[j] & (1 << i);
  long long k;
  for (int i = 0; i < n; i++) {
    k = 0;
    for (int j = 0; j <= 30; j++) {
      if (a[i] & (1 << j))
        k += n * (1 << j);
      else
        k += s[j];
    }
    if (k != c[i]) return cout << -1, 0;
    k = 0;
    for (int j = 0; j <= 30; j++)
      if (a[i] & (1 << j)) k += s[j];
    if (k != b[i]) return cout << -1, 0;
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
}

```