#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 2e5 + 4;
int main() {
  int n;
  scanf("%d", &n);
  long long b[n], c[n], sum = 0;
  for (int i = 0; i < n; i++) scanf("%lld", &b[i]), sum += b[i];
  for (int i = 0; i < n; i++) scanf("%lld", &c[i]), sum += c[i];
  if (sum % 2 * n != 0) {
    printf("%d", -1);
    return 0;
  }
  sum /= 2 * n;
  long long a[n];
  for (int i = 0; i < n; i++) {
    long long here = b[i] + c[i] - sum;
    if (here >= 0 and here % n != 0) {
      printf("%d", -1);
      return 0;
    }
    a[i] = here / n;
  }
  vector<int> fa(31, 0);
  for (int bit = 0; bit < 31; bit++) {
    for (int i = 0; i < n; i++)
      if (a[i] & (1LL << bit)) fa[bit]++;
  }
  for (int i = 0; i < n; i++) {
    for (int bit = 0; bit < 31; bit++)
      if (a[i] & (1LL << bit)) b[i] -= (1LL << bit) * fa[bit];
    if (b[i] != 0) {
      printf("%d", -1);
      return 0;
    }
  }
  for (int i = 0; i < n; i++) printf("%lld ", a[i]);
  return 0;
}

### Prompt

Create a solution in cpp for 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 long long mod = 1e9 + 7;
const int N = 2e5 + 4;
int main() {
  int n;
  scanf("%d", &n);
  long long b[n], c[n], sum = 0;
  for (int i = 0; i < n; i++) scanf("%lld", &b[i]), sum += b[i];
  for (int i = 0; i < n; i++) scanf("%lld", &c[i]), sum += c[i];
  if (sum % 2 * n != 0) {
    printf("%d", -1);
    return 0;
  }
  sum /= 2 * n;
  long long a[n];
  for (int i = 0; i < n; i++) {
    long long here = b[i] + c[i] - sum;
    if (here >= 0 and here % n != 0) {
      printf("%d", -1);
      return 0;
    }
    a[i] = here / n;
  }
  vector<int> fa(31, 0);
  for (int bit = 0; bit < 31; bit++) {
    for (int i = 0; i < n; i++)
      if (a[i] & (1LL << bit)) fa[bit]++;
  }
  for (int i = 0; i < n; i++) {
    for (int bit = 0; bit < 31; bit++)
      if (a[i] & (1LL << bit)) b[i] -= (1LL << bit) * fa[bit];
    if (b[i] != 0) {
      printf("%d", -1);
      return 0;
    }
  }
  for (int i = 0; i < n; i++) printf("%lld ", a[i]);
  return 0;
}

```