#include <bits/stdc++.h>
using namespace std;
long long n, bb[200005], cc[200005], sum, a[200005], b[200005], c[200005],
    cnt[32];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> bb[i], sum += bb[i];
  for (int i = 1; i <= n; ++i) cin >> cc[i], sum += cc[i];
  sum /= 2ll * n;
  for (int i = 1; i <= n; ++i) a[i] = (bb[i] + cc[i] - sum) / n;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 31; ++j)
      if (a[i] & (1ll << j)) ++cnt[j];
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 31; ++j)
      if (a[i] & (1ll << j)) b[i] += (1ll << j) * cnt[j];
    for (int j = 0; j <= 31; ++j)
      if (a[i] & (1ll << j))
        c[i] += (1ll << j) * n;
      else
        c[i] += (1ll << j) * cnt[j];
  }
  for (int i = 1; i <= n; ++i)
    if (b[i] != bb[i] || c[i] != cc[i]) {
      cout << -1;
      return 0;
    }
  for (int i = 1; i <= n; ++i) cout << a[i] << ' ';
  cout << endl;
  return 0;
}

### Prompt

Generate 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, bb[200005], cc[200005], sum, a[200005], b[200005], c[200005],
    cnt[32];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> bb[i], sum += bb[i];
  for (int i = 1; i <= n; ++i) cin >> cc[i], sum += cc[i];
  sum /= 2ll * n;
  for (int i = 1; i <= n; ++i) a[i] = (bb[i] + cc[i] - sum) / n;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 31; ++j)
      if (a[i] & (1ll << j)) ++cnt[j];
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 31; ++j)
      if (a[i] & (1ll << j)) b[i] += (1ll << j) * cnt[j];
    for (int j = 0; j <= 31; ++j)
      if (a[i] & (1ll << j))
        c[i] += (1ll << j) * n;
      else
        c[i] += (1ll << j) * cnt[j];
  }
  for (int i = 1; i <= n; ++i)
    if (b[i] != bb[i] || c[i] != cc[i]) {
      cout << -1;
      return 0;
    }
  for (int i = 1; i <= n; ++i) cout << a[i] << ' ';
  cout << endl;
  return 0;
}

```