#include <bits/stdc++.h>
using namespace std;
int frc[100];
int b[200010], c[200010], a[200010];
int main() {
  int n;
  scanf("%d", &n);
  long long s(0);
  for (int i(0); i < n; i++) scanf("%d", &b[i]);
  for (int i(0); i < n; i++) scanf("%d", &c[i]);
  for (int i(0); i < n; i++) s += 1ll * b[i] + 1ll * c[i];
  if (s % (2 * n) != 0) {
    cout << -1;
    return 0;
  }
  s /= 2 * n;
  for (int i(0); i < n; i++) {
    if ((b[i] + c[i] - s) % n != 0) {
      cout << -1;
      return 0;
    }
    a[i] = (b[i] + c[i] - s) / n;
    for (int j(0); j <= 30; j++)
      if ((a[i] & (1 << j))) frc[j]++;
  }
  for (int i(0); i < n; i++) {
    long long c0(0), b0(0);
    for (int j(0); j <= 30; j++) {
      if ((a[i] & (1 << j)))
        c0 += 1ll * n * (1 << j), b0 += 1ll * frc[j] * (1 << j);
      else
        c0 += 1ll * frc[j] * (1 << j);
    }
    if (c0 != c[i] || b0 != b[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i(0); i < n; i++) cout << a[i] << ' ';
  return 0;
}

### Prompt

Develop a solution in CPP 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;
int frc[100];
int b[200010], c[200010], a[200010];
int main() {
  int n;
  scanf("%d", &n);
  long long s(0);
  for (int i(0); i < n; i++) scanf("%d", &b[i]);
  for (int i(0); i < n; i++) scanf("%d", &c[i]);
  for (int i(0); i < n; i++) s += 1ll * b[i] + 1ll * c[i];
  if (s % (2 * n) != 0) {
    cout << -1;
    return 0;
  }
  s /= 2 * n;
  for (int i(0); i < n; i++) {
    if ((b[i] + c[i] - s) % n != 0) {
      cout << -1;
      return 0;
    }
    a[i] = (b[i] + c[i] - s) / n;
    for (int j(0); j <= 30; j++)
      if ((a[i] & (1 << j))) frc[j]++;
  }
  for (int i(0); i < n; i++) {
    long long c0(0), b0(0);
    for (int j(0); j <= 30; j++) {
      if ((a[i] & (1 << j)))
        c0 += 1ll * n * (1 << j), b0 += 1ll * frc[j] * (1 << j);
      else
        c0 += 1ll * frc[j] * (1 << j);
    }
    if (c0 != c[i] || b0 != b[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i(0); i < n; i++) cout << a[i] << ' ';
  return 0;
}

```