#include <bits/stdc++.h>
using namespace std;
const int N = 200001;
long long n, i, sum, B, C, j, k, a[N], b[N], c[N], d[N], s[30], bit[30];
void ns() {
  printf("-1");
  exit(0);
}
int main() {
  scanf("%I64d\n", &n);
  for (i = 1; i <= n; i++) scanf("%I64d", &b[i]);
  for (scanf("\n"), i = 1; i <= n; i++)
    scanf("%I64d", &c[i]), d[i] = b[i] + c[i], sum += d[i];
  if (sum % (2 * n)) ns();
  sum /= 2 * n;
  for (i = 1; i <= n; i++) {
    a[i] = d[i] - sum;
    if (a[i] % n) ns();
    a[i] /= n;
  }
  for (s[0] = i = 1; i <= 29; i++) s[i] = s[i - 1] << 1;
  for (i = 1; i <= n; i++)
    for (j = a[i], k = 0; k <= 29; k++, j >>= 1) bit[k] += j & 1;
  for (i = 1; i <= n; i++) {
    B = C = 0;
    for (j = a[i], k = 0; k <= 29; k++, j >>= 1)
      if (j & 1)
        B += s[k] * bit[k], C += s[k] * n;
      else
        C += s[k] * bit[k];
    if (B != b[i] || C != c[i]) ns();
  }
  for (i = 1; i <= n; i++) printf("%I64d ", a[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;
const int N = 200001;
long long n, i, sum, B, C, j, k, a[N], b[N], c[N], d[N], s[30], bit[30];
void ns() {
  printf("-1");
  exit(0);
}
int main() {
  scanf("%I64d\n", &n);
  for (i = 1; i <= n; i++) scanf("%I64d", &b[i]);
  for (scanf("\n"), i = 1; i <= n; i++)
    scanf("%I64d", &c[i]), d[i] = b[i] + c[i], sum += d[i];
  if (sum % (2 * n)) ns();
  sum /= 2 * n;
  for (i = 1; i <= n; i++) {
    a[i] = d[i] - sum;
    if (a[i] % n) ns();
    a[i] /= n;
  }
  for (s[0] = i = 1; i <= 29; i++) s[i] = s[i - 1] << 1;
  for (i = 1; i <= n; i++)
    for (j = a[i], k = 0; k <= 29; k++, j >>= 1) bit[k] += j & 1;
  for (i = 1; i <= n; i++) {
    B = C = 0;
    for (j = a[i], k = 0; k <= 29; k++, j >>= 1)
      if (j & 1)
        B += s[k] * bit[k], C += s[k] * n;
      else
        C += s[k] * bit[k];
    if (B != b[i] || C != c[i]) ns();
  }
  for (i = 1; i <= n; i++) printf("%I64d ", a[i]);
}

```