#include <bits/stdc++.h>
using namespace std;
long long int a[200001], b[200001], c[200001], cnt[45][2];
int main() {
  ios::sync_with_stdio(false);
  ;
  cin.tie(0);
  ;
  long long int i, j, n, sum = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> b[i];
    sum += b[i];
  }
  for (i = 1; i <= n; i++) {
    cin >> c[i];
    sum += c[i];
  }
  if (sum % (2 * n)) {
    cout << "-1";
    return 0;
  }
  sum = sum / (2 * n);
  for (i = 1; i <= n; i++) {
    if ((b[i] + c[i] - sum) % n || b[i] + c[i] - sum < 0) {
      cout << "-1";
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j < 40; j++) {
      if (a[i] & (1LL << j))
        cnt[j][1]++;
      else
        cnt[j][0]++;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j < 40; j++) {
      if (a[i] & (1LL << j)) {
        b[i] = b[i] - (1LL << j) * cnt[j][1];
        c[i] = c[i] - (1LL << j) * (cnt[j][0] + cnt[j][1]);
      } else
        c[i] = c[i] - (1LL << j) * cnt[j][1];
    }
    if (b[i] != 0 || c[i] != 0) {
      cout << "-1";
      return 0;
    }
  }
  for (i = 1; i <= n; i++) cout << a[i] << " ";
  return 0;
}

### Prompt

In CPP, your task is to solve 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 int a[200001], b[200001], c[200001], cnt[45][2];
int main() {
  ios::sync_with_stdio(false);
  ;
  cin.tie(0);
  ;
  long long int i, j, n, sum = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> b[i];
    sum += b[i];
  }
  for (i = 1; i <= n; i++) {
    cin >> c[i];
    sum += c[i];
  }
  if (sum % (2 * n)) {
    cout << "-1";
    return 0;
  }
  sum = sum / (2 * n);
  for (i = 1; i <= n; i++) {
    if ((b[i] + c[i] - sum) % n || b[i] + c[i] - sum < 0) {
      cout << "-1";
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j < 40; j++) {
      if (a[i] & (1LL << j))
        cnt[j][1]++;
      else
        cnt[j][0]++;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j < 40; j++) {
      if (a[i] & (1LL << j)) {
        b[i] = b[i] - (1LL << j) * cnt[j][1];
        c[i] = c[i] - (1LL << j) * (cnt[j][0] + cnt[j][1]);
      } else
        c[i] = c[i] - (1LL << j) * cnt[j][1];
    }
    if (b[i] != 0 || c[i] != 0) {
      cout << "-1";
      return 0;
    }
  }
  for (i = 1; i <= n; i++) cout << a[i] << " ";
  return 0;
}

```