#include <bits/stdc++.h>
using namespace std;
bool solve() {
  int n;
  cin >> n;
  vector<int64_t> b(n), c(n);
  int64_t total = 0;
  for (int i = 0; i < n; i++) scanf("%lld", &b[i]), total += b[i];
  for (int i = 0; i < n; i++) scanf("%lld", &c[i]), total += c[i];
  if (total % (2 * n) != 0) return false;
  int64_t sum = total / 2 / n;
  vector<int64_t> a(n);
  for (int i = 0; i < n; i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] < 0 || a[i] % n != 0) return false;
    a[i] /= n;
  }
  vector<int64_t> x(60);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 60; j++) {
      x[j] += a[i] >> j & 1;
    }
  }
  for (int i = 0; i < n; i++) {
    int64_t xx = 0, yy = 0;
    for (int j = 0; j < 60; j++) {
      if (a[i] >> j & 1) {
        xx += (1LL << j) * x[j];
        yy += (1LL << j) * n;
      } else {
        yy += (1LL << j) * x[j];
      }
    }
    if (xx != b[i] || yy != c[i]) return false;
  }
  for (int i = 0; i < n; i++) printf("%lld ", a[i]);
  return true;
}
int main() {
  if (!solve()) puts("-1");
}

### 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;
bool solve() {
  int n;
  cin >> n;
  vector<int64_t> b(n), c(n);
  int64_t total = 0;
  for (int i = 0; i < n; i++) scanf("%lld", &b[i]), total += b[i];
  for (int i = 0; i < n; i++) scanf("%lld", &c[i]), total += c[i];
  if (total % (2 * n) != 0) return false;
  int64_t sum = total / 2 / n;
  vector<int64_t> a(n);
  for (int i = 0; i < n; i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] < 0 || a[i] % n != 0) return false;
    a[i] /= n;
  }
  vector<int64_t> x(60);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 60; j++) {
      x[j] += a[i] >> j & 1;
    }
  }
  for (int i = 0; i < n; i++) {
    int64_t xx = 0, yy = 0;
    for (int j = 0; j < 60; j++) {
      if (a[i] >> j & 1) {
        xx += (1LL << j) * x[j];
        yy += (1LL << j) * n;
      } else {
        yy += (1LL << j) * x[j];
      }
    }
    if (xx != b[i] || yy != c[i]) return false;
  }
  for (int i = 0; i < n; i++) printf("%lld ", a[i]);
  return true;
}
int main() {
  if (!solve()) puts("-1");
}

```