#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 200005;
int n, a[N], b[N], c[N], s[N], cnt[30];
bool check() {
  for (int i = 1; i <= n; i++) {
    ll bb = 0, cc = 0;
    for (int j = 0; j < 30; j++) {
      if (a[i] & (1 << j))
        bb += (1 << j) * cnt[j], cc += (1 << j) * n;
      else
        cc += (1 << j) * cnt[j];
    }
    if (bb != b[i] || cc != c[i]) return 0;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) cin >> c[i];
  ll sum = 0;
  for (int i = 1; i <= n; i++) s[i] = b[i] + c[i], sum += s[i];
  sum /= n * 2;
  for (int i = 1; i <= n; i++) {
    a[i] = (s[i] - sum) / n;
    for (int j = 0; j < 30; j++)
      if (a[i] & (1 << j)) ++cnt[j];
  }
  if (!check())
    cout << -1 << endl;
  else
    for (int i = 1; i <= n; i++) cout << a[i] << ' ';
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
using ll = long long;
const int N = 200005;
int n, a[N], b[N], c[N], s[N], cnt[30];
bool check() {
  for (int i = 1; i <= n; i++) {
    ll bb = 0, cc = 0;
    for (int j = 0; j < 30; j++) {
      if (a[i] & (1 << j))
        bb += (1 << j) * cnt[j], cc += (1 << j) * n;
      else
        cc += (1 << j) * cnt[j];
    }
    if (bb != b[i] || cc != c[i]) return 0;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) cin >> c[i];
  ll sum = 0;
  for (int i = 1; i <= n; i++) s[i] = b[i] + c[i], sum += s[i];
  sum /= n * 2;
  for (int i = 1; i <= n; i++) {
    a[i] = (s[i] - sum) / n;
    for (int j = 0; j < 30; j++)
      if (a[i] & (1 << j)) ++cnt[j];
  }
  if (!check())
    cout << -1 << endl;
  else
    for (int i = 1; i <= n; i++) cout << a[i] << ' ';
  return 0;
}

```