#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 10;
long long a[maxn], b[maxn], c[maxn];
int num[64];
int n;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    a[i] = b[i] + c[i];
    sum += a[i];
  }
  if (sum % (2 * n)) {
    cout << -1 << endl;
    return 0;
  }
  sum /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = (a[i] - sum);
    if (a[i] % n) {
      cout << -1 << endl;
      return 0;
    } else {
      a[i] /= n;
    }
  }
  for (int i = 0; i < n; i++) {
    long long temp = a[i];
    for (int j = 0; j < 64; j++) {
      num[j] += temp % 2;
      temp /= 2;
    }
  }
  for (int i = 0; i < n; i++) {
    long long b_ = 0, c_ = 0;
    for (int j = 0; j < 64; j++) {
      if (a[i] & (1LL << j)) {
        b_ += num[j] * (1LL << j);
        c_ += n * (1LL << j);
      } else
        c_ += num[j] * (1LL << j);
    }
    if (b_ != b[i] || c_ != c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << a[0];
  for (int i = 1; i < n; i++) {
    cout << " " << a[i];
  }
  cout << endl;
  return 0;
}

### 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 int maxn = 200000 + 10;
long long a[maxn], b[maxn], c[maxn];
int num[64];
int n;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    a[i] = b[i] + c[i];
    sum += a[i];
  }
  if (sum % (2 * n)) {
    cout << -1 << endl;
    return 0;
  }
  sum /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = (a[i] - sum);
    if (a[i] % n) {
      cout << -1 << endl;
      return 0;
    } else {
      a[i] /= n;
    }
  }
  for (int i = 0; i < n; i++) {
    long long temp = a[i];
    for (int j = 0; j < 64; j++) {
      num[j] += temp % 2;
      temp /= 2;
    }
  }
  for (int i = 0; i < n; i++) {
    long long b_ = 0, c_ = 0;
    for (int j = 0; j < 64; j++) {
      if (a[i] & (1LL << j)) {
        b_ += num[j] * (1LL << j);
        c_ += n * (1LL << j);
      } else
        c_ += num[j] * (1LL << j);
    }
    if (b_ != b[i] || c_ != c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << a[0];
  for (int i = 1; i < n; i++) {
    cout << " " << a[i];
  }
  cout << endl;
  return 0;
}

```