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

### Prompt

Your challenge is to write 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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n), b(n), c(n);
  long long 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)) return cout << -1, 0;
  sum /= 2 * n;
  for (int i = 0; i < n; i++) {
    if ((b[i] + c[i] - sum) % n) return cout << -1, 0;
    a[i] = (b[i] + c[i] - sum) / n;
  }
  vector<long long> cou(30);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) cou[j]++;
    }
  }
  vector<long long> tb(n), tc(n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) {
        tb[i] += (1 << j) * cou[j];
        tc[i] += (1 << j) * n;
      } else {
        tc[i] += (1 << j) * cou[j];
      }
    }
  }
  if (b != tb || c != tc) return cout << -1, 0;
  for (long long x : a) cout << x << " ";
}

```