#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int a[N];
int b[N];
int c[N];
long long sum_bc;
long long sum_a;
int cnt[32];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
    sum_bc += b[i] + c[i];
  }
  if (sum_bc % (2 * n)) {
    cout << -1;
    return 0;
  }
  sum_a = sum_bc / (2 * n);
  for (int i = 0; i < n; ++i) {
    a[i] = (c[i] + b[i] - sum_a) / n;
    if (a[i] < 0) {
      cout << -1;
      return 0;
    }
    for (int j = 0; j < 32; ++j) {
      cnt[j] += (a[i] >> j) & 1;
    }
  }
  for (int i = 0; i < n; ++i) {
    long long res_b = 0;
    for (int j = 0; j < 32; ++j) {
      if ((a[i] >> j) & 1) {
        res_b += (1 << j) * cnt[j];
      }
    }
    if (res_b != b[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << a[i] << ' ';
  }
  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 N = 200005;
int a[N];
int b[N];
int c[N];
long long sum_bc;
long long sum_a;
int cnt[32];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
    sum_bc += b[i] + c[i];
  }
  if (sum_bc % (2 * n)) {
    cout << -1;
    return 0;
  }
  sum_a = sum_bc / (2 * n);
  for (int i = 0; i < n; ++i) {
    a[i] = (c[i] + b[i] - sum_a) / n;
    if (a[i] < 0) {
      cout << -1;
      return 0;
    }
    for (int j = 0; j < 32; ++j) {
      cnt[j] += (a[i] >> j) & 1;
    }
  }
  for (int i = 0; i < n; ++i) {
    long long res_b = 0;
    for (int j = 0; j < 32; ++j) {
      if ((a[i] >> j) & 1) {
        res_b += (1 << j) * cnt[j];
      }
    }
    if (res_b != b[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << a[i] << ' ';
  }
  return 0;
}

```