#include <bits/stdc++.h>
using namespace std;
int n;
long long v[200010], a[200010], b[200010], c[200010], dif[200010];
bool check() {
  for (int bit = 0; bit < 30; ++bit) {
    long long sum = 0;
    for (int i = 1; i <= n; ++i) {
      if ((v[i] >> bit) & 1) {
        sum += (1 << bit);
      }
    }
    for (int i = 1; i <= n; ++i) {
      if ((v[i] >> bit) & 1) {
        b[i] -= sum;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (b[i] != 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> b[i];
  }
  for (int i = 1; i <= n; ++i) {
    cin >> c[i];
  }
  for (int i = 1; i <= n; ++i) {
    a[i] = b[i] + c[i];
  }
  long long sum = 0;
  for (int i = 2; i <= n; ++i) {
    dif[i] = (a[i] - a[i - 1]) / n + dif[i - 1];
    sum += dif[i];
  }
  v[1] = (a[1] - sum) / (2 * n);
  for (int i = 2; i <= n; ++i) {
    v[i] = v[1] + dif[i];
  }
  if (!check()) {
    cout << -1;
  } else {
    for (int i = 1; i <= n; ++i) {
      cout << v[i] << " ";
    }
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 n;
long long v[200010], a[200010], b[200010], c[200010], dif[200010];
bool check() {
  for (int bit = 0; bit < 30; ++bit) {
    long long sum = 0;
    for (int i = 1; i <= n; ++i) {
      if ((v[i] >> bit) & 1) {
        sum += (1 << bit);
      }
    }
    for (int i = 1; i <= n; ++i) {
      if ((v[i] >> bit) & 1) {
        b[i] -= sum;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (b[i] != 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> b[i];
  }
  for (int i = 1; i <= n; ++i) {
    cin >> c[i];
  }
  for (int i = 1; i <= n; ++i) {
    a[i] = b[i] + c[i];
  }
  long long sum = 0;
  for (int i = 2; i <= n; ++i) {
    dif[i] = (a[i] - a[i - 1]) / n + dif[i - 1];
    sum += dif[i];
  }
  v[1] = (a[1] - sum) / (2 * n);
  for (int i = 2; i <= n; ++i) {
    v[i] = v[1] + dif[i];
  }
  if (!check()) {
    cout << -1;
  } else {
    for (int i = 1; i <= n; ++i) {
      cout << v[i] << " ";
    }
  }
}

```