#include <bits/stdc++.h>
using namespace std;
long long n, b[(200000 + 7)], c[(200000 + 7)], z[(200000 + 7)], sum,
    a[(200000 + 7)], b2[(200000 + 7)], c2[(200000 + 7)];
int main() {
  scanf("%I64d", &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) {
    if (b[i] > c[i]) {
      printf("-1\n");
      return 0;
    }
    z[i] = b[i] + c[i];
    sum += z[i];
  }
  sum /= 2;
  sum /= n;
  for (int i = 1; i <= n; ++i) a[i] = (z[i] - sum) / n;
  for (int pw = 31; pw >= 0; --pw) {
    int cnt = 0;
    for (int i = 1; i <= n; ++i)
      if ((1 << pw) & a[i]) cnt++;
    for (int i = 1; i <= n; ++i) {
      if ((1 << pw) & a[i]) {
        b2[i] += cnt * (1 << pw);
        c2[i] += n * (1 << pw);
      } else {
        c2[i] += cnt * (1 << pw);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (b[i] != b2[i] || c[i] != c2[i]) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; ++i) printf("%I64d ", a[i]);
  printf("\n");
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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, b[(200000 + 7)], c[(200000 + 7)], z[(200000 + 7)], sum,
    a[(200000 + 7)], b2[(200000 + 7)], c2[(200000 + 7)];
int main() {
  scanf("%I64d", &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) {
    if (b[i] > c[i]) {
      printf("-1\n");
      return 0;
    }
    z[i] = b[i] + c[i];
    sum += z[i];
  }
  sum /= 2;
  sum /= n;
  for (int i = 1; i <= n; ++i) a[i] = (z[i] - sum) / n;
  for (int pw = 31; pw >= 0; --pw) {
    int cnt = 0;
    for (int i = 1; i <= n; ++i)
      if ((1 << pw) & a[i]) cnt++;
    for (int i = 1; i <= n; ++i) {
      if ((1 << pw) & a[i]) {
        b2[i] += cnt * (1 << pw);
        c2[i] += n * (1 << pw);
      } else {
        c2[i] += cnt * (1 << pw);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (b[i] != b2[i] || c[i] != c2[i]) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; ++i) printf("%I64d ", a[i]);
  printf("\n");
  return 0;
}

```