#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5, inf = 1e18;
long long n, a[N], b[N], c[N], t[41], ps;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) cin >> c[i], ps += c[i] + b[i];
  long long second = ps / (2 * n);
  for (long long i = 1; i <= n; i++) {
    a[i] = (b[i] + c[i] - second) / n;
    for (long long j = 40; j >= 0; j--)
      if ((a[i] & (1ll << j))) t[j]++;
  }
  for (long long i = 1; i <= n; i++) {
    long long B = 0, C = 0;
    for (long long j = 40; j >= 0; j--) {
      if ((a[i] & (1ll << j))) B += (1ll << j) * t[j];
      if ((a[i] & (1ll << j)))
        C += n * (1ll << j);
      else
        C += t[j] * (1ll << j);
    }
    if (b[i] != B || c[i] != C) return cout << -1, 0;
  }
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
  return 0;
}

### 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;
const long long N = 2e5 + 5, inf = 1e18;
long long n, a[N], b[N], c[N], t[41], ps;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) cin >> c[i], ps += c[i] + b[i];
  long long second = ps / (2 * n);
  for (long long i = 1; i <= n; i++) {
    a[i] = (b[i] + c[i] - second) / n;
    for (long long j = 40; j >= 0; j--)
      if ((a[i] & (1ll << j))) t[j]++;
  }
  for (long long i = 1; i <= n; i++) {
    long long B = 0, C = 0;
    for (long long j = 40; j >= 0; j--) {
      if ((a[i] & (1ll << j))) B += (1ll << j) * t[j];
      if ((a[i] & (1ll << j)))
        C += n * (1ll << j);
      else
        C += t[j] * (1ll << j);
    }
    if (b[i] != B || c[i] != C) return cout << -1, 0;
  }
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
  return 0;
}

```