#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
long long n, sum = 0;
long long a[N], b[N], c[N];
long long ss[30];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  bool flag = true;
  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];
    if (c[i] < b[i]) flag = false;
  }
  if (sum % (2 * n)) flag = false;
  sum /= 2 * n;
  for (int i = 0; i < n; ++i) {
    if ((b[i] + c[i] - sum) % n) flag = false;
    a[i] = (b[i] + c[i] - sum) / n;
  }
  memset(ss, 0, sizeof ss);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 30; ++j) {
      if (a[i] & (1 << j)) {
        ss[j] += (1 << j);
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    long long tmp = 0;
    for (int j = 0; j < 30; ++j) {
      if (a[i] & (1 << j)) {
        tmp += ss[j];
      }
    }
    if (tmp != b[i]) flag = false;
  }
  if (!flag)
    cout << -1 << '\n';
  else {
    for (int i = 0; i < n; ++i) cout << a[i] << " ";
    cout << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 N = 2e5 + 10;
long long n, sum = 0;
long long a[N], b[N], c[N];
long long ss[30];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  bool flag = true;
  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];
    if (c[i] < b[i]) flag = false;
  }
  if (sum % (2 * n)) flag = false;
  sum /= 2 * n;
  for (int i = 0; i < n; ++i) {
    if ((b[i] + c[i] - sum) % n) flag = false;
    a[i] = (b[i] + c[i] - sum) / n;
  }
  memset(ss, 0, sizeof ss);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 30; ++j) {
      if (a[i] & (1 << j)) {
        ss[j] += (1 << j);
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    long long tmp = 0;
    for (int j = 0; j < 30; ++j) {
      if (a[i] & (1 << j)) {
        tmp += ss[j];
      }
    }
    if (tmp != b[i]) flag = false;
  }
  if (!flag)
    cout << -1 << '\n';
  else {
    for (int i = 0; i < n; ++i) cout << a[i] << " ";
    cout << '\n';
  }
  return 0;
}

```