#include <bits/stdc++.h>
using namespace std;
const long long MAX = 2e5 + 9, LG = 35;
long long sum, b[MAX], c[MAX], a[MAX], n, k[LG];
int main() {
  cin >> n;
  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) != 0) return cout << -1, 0;
  sum /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] % n != 0 || a[i] < 0) return cout << -1, 0;
    a[i] /= n;
    long long x = a[i];
    for (int j = 0; j < LG; j++) k[j] += x >> j & 1;
  }
  for (int i = 0; i < n; i++) {
    long long sum1 = 0, sum2 = 0;
    for (int j = 0; j < LG; j++)
      if (a[i] >> j & 1)
        sum1 += k[j] * (1 << j), sum2 += n * (1 << j);
      else
        sum2 += k[j] * (1 << j);
    if (sum1 != b[i] || sum2 != c[i]) return cout << -1, 0;
  }
  for (int i = 0; i < n; i++) cout << a[i] << ' ';
}

### Prompt

Your task is to create 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 MAX = 2e5 + 9, LG = 35;
long long sum, b[MAX], c[MAX], a[MAX], n, k[LG];
int main() {
  cin >> n;
  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) != 0) return cout << -1, 0;
  sum /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] % n != 0 || a[i] < 0) return cout << -1, 0;
    a[i] /= n;
    long long x = a[i];
    for (int j = 0; j < LG; j++) k[j] += x >> j & 1;
  }
  for (int i = 0; i < n; i++) {
    long long sum1 = 0, sum2 = 0;
    for (int j = 0; j < LG; j++)
      if (a[i] >> j & 1)
        sum1 += k[j] * (1 << j), sum2 += n * (1 << j);
      else
        sum2 += k[j] * (1 << j);
    if (sum1 != b[i] || sum2 != c[i]) return cout << -1, 0;
  }
  for (int i = 0; i < n; i++) cout << a[i] << ' ';
}

```