#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL n, a[200 * 1000], d[200 * 1000], b[200 * 1000], c[200 * 1000];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  d[0] = 0;
  LL sumd = 0;
  for (int i = 1; i < n; i++) {
    LL tmp = (b[i] + c[i]) - (b[i - 1] + c[i - 1]);
    d[i] = d[i - 1] + tmp / n;
    sumd += d[i];
  }
  LL a0 = (b[0] + c[0] - sumd) / (n * 2);
  for (int i = 0; i < n; i++) a[i] = a0 + d[i];
  bool fail = false;
  for (int i = 0; i < n; i++)
    if (a[i] < 0) fail = true;
  for (int d = 1 << 30; d; d /= 2) {
    LL cnt = 0;
    for (int i = 0; i < n; i++)
      if (a[i] & d) cnt++;
    for (int i = 0; i < n; i++) {
      if (a[i] & d)
        b[i] -= cnt * d;
      else
        c[i] += (n - cnt) * d;
    }
  }
  LL full = (1LL << 31) - 1;
  for (int i = 0; i < n; i++) {
    if (b[i] != 0 || c[i] != full * n) fail = true;
  }
  if (fail) {
    cout << -1 << endl;
  } else {
    for (int i = 0; i < n; i++) cout << a0 + d[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;
typedef long long LL;
LL n, a[200 * 1000], d[200 * 1000], b[200 * 1000], c[200 * 1000];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  d[0] = 0;
  LL sumd = 0;
  for (int i = 1; i < n; i++) {
    LL tmp = (b[i] + c[i]) - (b[i - 1] + c[i - 1]);
    d[i] = d[i - 1] + tmp / n;
    sumd += d[i];
  }
  LL a0 = (b[0] + c[0] - sumd) / (n * 2);
  for (int i = 0; i < n; i++) a[i] = a0 + d[i];
  bool fail = false;
  for (int i = 0; i < n; i++)
    if (a[i] < 0) fail = true;
  for (int d = 1 << 30; d; d /= 2) {
    LL cnt = 0;
    for (int i = 0; i < n; i++)
      if (a[i] & d) cnt++;
    for (int i = 0; i < n; i++) {
      if (a[i] & d)
        b[i] -= cnt * d;
      else
        c[i] += (n - cnt) * d;
    }
  }
  LL full = (1LL << 31) - 1;
  for (int i = 0; i < n; i++) {
    if (b[i] != 0 || c[i] != full * n) fail = true;
  }
  if (fail) {
    cout << -1 << endl;
  } else {
    for (int i = 0; i < n; i++) cout << a0 + d[i] << ' ';
    cout << endl;
  }
  return 0;
}

```