#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
long long a[N], b[N], c[N], cnt[60];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]);
  for (int i = 1; i <= n; i++) scanf("%I64d", &c[i]);
  long long s = 0;
  for (int i = 1; i <= n; i++) s += b[i] + c[i];
  if (s % (2 * n)) return 0 * printf("-1");
  s /= 2 * n;
  for (int i = 1; i <= n; i++) {
    if (b[i] + c[i] - s < 0 || (b[i] + c[i] - s) % n) return 0 * printf("-1");
    a[i] = (b[i] + c[i] - s) / n;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 60; j++) cnt[j] += (a[i] >> j & 1);
  for (int i = 1; i <= n; i++) {
    long long tb = 0, tc = 0;
    for (int j = 0; j < 60; j++)
      if ((a[i] >> j & 1)) {
        tb += cnt[j] * (1LL << j);
        tc += n * (1LL << j);
      } else
        tc += cnt[j] * (1LL << j);
    if (tb != b[i] || tc != c[i]) return 0 * printf("-1");
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}

### 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;
const int N = 200005;
long long a[N], b[N], c[N], cnt[60];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]);
  for (int i = 1; i <= n; i++) scanf("%I64d", &c[i]);
  long long s = 0;
  for (int i = 1; i <= n; i++) s += b[i] + c[i];
  if (s % (2 * n)) return 0 * printf("-1");
  s /= 2 * n;
  for (int i = 1; i <= n; i++) {
    if (b[i] + c[i] - s < 0 || (b[i] + c[i] - s) % n) return 0 * printf("-1");
    a[i] = (b[i] + c[i] - s) / n;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 60; j++) cnt[j] += (a[i] >> j & 1);
  for (int i = 1; i <= n; i++) {
    long long tb = 0, tc = 0;
    for (int j = 0; j < 60; j++)
      if ((a[i] >> j & 1)) {
        tb += cnt[j] * (1LL << j);
        tc += n * (1LL << j);
      } else
        tc += cnt[j] * (1LL << j);
    if (tb != b[i] || tc != c[i]) return 0 * printf("-1");
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}

```