#include <bits/stdc++.h>
using namespace std;
long long a[200009], b[200009], c[200009], k[64];
const long long one = 1;
int main() {
  int n;
  cin >> n;
  long long sum = 0;
  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];
  }
  int ans = 0;
  if (sum % (2 * n) != 0)
    ans = -1;
  else
    sum /= (2 * n);
  for (int i = 0; i < n; ++i) {
    if (ans == -1 || b[i] + c[i] < sum || (b[i] + c[i] - sum) % n != 0) {
      ans = -1;
      break;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  memset(k, 0, sizeof(k));
  for (int j = 0; j < 32; ++j) {
    for (int i = 0; i < n; ++i) {
      k[j] += ((a[i] >> j) & one);
    }
  }
  for (int i = 0; i < n; ++i) {
    long long bi = 0;
    long long ci = 0;
    for (int j = 0; j < 32; ++j) {
      if ((a[i] >> j) & one) {
        bi += k[j] * (one << j);
        ci += n * (one << j);
      } else {
        ci += k[j] * (1 << j);
      }
    }
    if (b[i] != bi || c[i] != ci) {
      ans = -1;
      break;
    }
  }
  if (ans == -1)
    cout << "-1" << endl;
  else {
    for (int i = 0; i < n; ++i) cout << a[i] << " ";
  }
  cout << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long a[200009], b[200009], c[200009], k[64];
const long long one = 1;
int main() {
  int n;
  cin >> n;
  long long sum = 0;
  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];
  }
  int ans = 0;
  if (sum % (2 * n) != 0)
    ans = -1;
  else
    sum /= (2 * n);
  for (int i = 0; i < n; ++i) {
    if (ans == -1 || b[i] + c[i] < sum || (b[i] + c[i] - sum) % n != 0) {
      ans = -1;
      break;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  memset(k, 0, sizeof(k));
  for (int j = 0; j < 32; ++j) {
    for (int i = 0; i < n; ++i) {
      k[j] += ((a[i] >> j) & one);
    }
  }
  for (int i = 0; i < n; ++i) {
    long long bi = 0;
    long long ci = 0;
    for (int j = 0; j < 32; ++j) {
      if ((a[i] >> j) & one) {
        bi += k[j] * (one << j);
        ci += n * (one << j);
      } else {
        ci += k[j] * (1 << j);
      }
    }
    if (b[i] != bi || c[i] != ci) {
      ans = -1;
      break;
    }
  }
  if (ans == -1)
    cout << "-1" << endl;
  else {
    for (int i = 0; i < n; ++i) cout << a[i] << " ";
  }
  cout << endl;
  return 0;
}

```