#include <bits/stdc++.h>
using namespace std;
long long n, a[1000000], sum, ans[1000000], p[1000000], q[1000000], s[1000000];
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), p[i] = a[i];
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    q[i] = x;
    a[i] += x;
    sum += a[i];
  }
  if (sum % (((long long)(2)) * n)) {
    puts("-1");
    return 0;
  }
  sum /= (((long long)(2)) * n);
  for (int i = 1; i <= n; i++) {
    a[i] -= sum;
    if (a[i] < 0 || a[i] % n) {
      puts("-1");
      return 0;
    }
    ans[i] = a[i] / n;
  }
  for (int i = 1; i <= n; i++)
    for (long long k = ans[i], j = 0; k; j++, k >>= 1) s[j] += k & 1;
  for (int i = 1; i <= n; i++) {
    long long x = 0, y = 0;
    for (long long k = ans[i], j = 0; j < 32; j++, k >>= 1)
      if (k & 1)
        x += n * (1 << j), y += s[j] * (1 << j);
      else
        x += s[j] * (1 << j);
    if (x != q[i] || y != p[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
}

### 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;
long long n, a[1000000], sum, ans[1000000], p[1000000], q[1000000], s[1000000];
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), p[i] = a[i];
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    q[i] = x;
    a[i] += x;
    sum += a[i];
  }
  if (sum % (((long long)(2)) * n)) {
    puts("-1");
    return 0;
  }
  sum /= (((long long)(2)) * n);
  for (int i = 1; i <= n; i++) {
    a[i] -= sum;
    if (a[i] < 0 || a[i] % n) {
      puts("-1");
      return 0;
    }
    ans[i] = a[i] / n;
  }
  for (int i = 1; i <= n; i++)
    for (long long k = ans[i], j = 0; k; j++, k >>= 1) s[j] += k & 1;
  for (int i = 1; i <= n; i++) {
    long long x = 0, y = 0;
    for (long long k = ans[i], j = 0; j < 32; j++, k >>= 1)
      if (k & 1)
        x += n * (1 << j), y += s[j] * (1 << j);
      else
        x += s[j] * (1 << j);
    if (x != q[i] || y != p[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
}

```