#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
long long a[N], b[N], c[N], d[N];
int n;
bool check() {
  for (int i = 1; i <= n; i++)
    if (a[i] < 0) return false;
  for (int k = 0; k <= 30; k++) {
    long long cnt = 0;
    for (int i = 1; i <= n; i++) cnt += a[i] >> k & 1;
    for (int i = 1; i <= n; i++) {
      if (a[i] >> k & 1) {
        b[i] -= cnt << k;
        c[i] -= n << k;
      } else {
        b[i] -= 0;
        c[i] -= cnt << k;
      }
    }
  }
  for (int i = 1; i <= n; i++)
    if (b[i] != 0) return false;
  for (int i = 1; i <= n; i++)
    if (c[i] != 0) return false;
  return true;
}
int main() {
  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);
  for (int i = 1; i <= n; i++) d[i] = b[i] + c[i];
  long long sum = 0;
  for (int i = 1; i <= n; i++) sum += d[i];
  sum /= 2 * n;
  for (int i = 1; i <= n; i++) a[i] = (d[i] - sum) / n;
  if (check()) {
    for (int i = 1; i <= n; i++) printf("%I64d%c", a[i], " \n"[i == n]);
  } else
    puts("-1");
  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 a[N], b[N], c[N], d[N];
int n;
bool check() {
  for (int i = 1; i <= n; i++)
    if (a[i] < 0) return false;
  for (int k = 0; k <= 30; k++) {
    long long cnt = 0;
    for (int i = 1; i <= n; i++) cnt += a[i] >> k & 1;
    for (int i = 1; i <= n; i++) {
      if (a[i] >> k & 1) {
        b[i] -= cnt << k;
        c[i] -= n << k;
      } else {
        b[i] -= 0;
        c[i] -= cnt << k;
      }
    }
  }
  for (int i = 1; i <= n; i++)
    if (b[i] != 0) return false;
  for (int i = 1; i <= n; i++)
    if (c[i] != 0) return false;
  return true;
}
int main() {
  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);
  for (int i = 1; i <= n; i++) d[i] = b[i] + c[i];
  long long sum = 0;
  for (int i = 1; i <= n; i++) sum += d[i];
  sum /= 2 * n;
  for (int i = 1; i <= n; i++) a[i] = (d[i] - sum) / n;
  if (check()) {
    for (int i = 1; i <= n; i++) printf("%I64d%c", a[i], " \n"[i == n]);
  } else
    puts("-1");
  return 0;
}

```