#include <bits/stdc++.h>
using namespace std;
const int M = 200005;
long long a[M], b[M], c[M], S, n;
int ct[34];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]), S += b[i];
  for (int i = 1; i <= n; i++) scanf("%I64d", &c[i]), S += c[i];
  if (S % (2LL * n)) {
    cout << "-1" << endl;
    return 0;
  }
  S /= 2LL * n;
  for (int i = 1; i <= n; i++)
    if ((b[i] + c[i] - S) % n) {
      cout << "-1" << endl;
      return 0;
    } else
      a[i] = (b[i] + c[i] - S) / n;
  for (int j = 0; j < 34; j++)
    for (int i = 1; i <= n; i++) ct[j] += (a[i] >> j) & 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 34; j++) b[i] -= (a[i] & (1ll << j)) * ct[j];
    if (b[i] != 0) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}

### Prompt

Please formulate 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 int M = 200005;
long long a[M], b[M], c[M], S, n;
int ct[34];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]), S += b[i];
  for (int i = 1; i <= n; i++) scanf("%I64d", &c[i]), S += c[i];
  if (S % (2LL * n)) {
    cout << "-1" << endl;
    return 0;
  }
  S /= 2LL * n;
  for (int i = 1; i <= n; i++)
    if ((b[i] + c[i] - S) % n) {
      cout << "-1" << endl;
      return 0;
    } else
      a[i] = (b[i] + c[i] - S) / n;
  for (int j = 0; j < 34; j++)
    for (int i = 1; i <= n; i++) ct[j] += (a[i] >> j) & 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 34; j++) b[i] -= (a[i] & (1ll << j)) * ct[j];
    if (b[i] != 0) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}

```