#include <bits/stdc++.h>
using namespace std;
long long n, b[200005], c[200005], sum = 0;
long long ssum[40];
long long L = 1;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &b[i]);
    sum += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
    sum += c[i];
  }
  if (sum % (2 * n)) {
    cout << -1;
    return 0;
  }
  sum /= 2 * n;
  for (int i = 1; i <= n; i++)
    if ((b[i] + c[i] - sum) % n || (b[i] + c[i] - sum) / n > b[i] ||
        (b[i] + c[i] - sum) / n > c[i]) {
      cout << -1;
      return 0;
    }
  long long x = 0;
  for (int i = 1; i <= n; i++)
    if (c[i] < b[i]) {
      cout << -1;
      return 0;
    }
  memset(ssum, 0, sizeof(ssum));
  for (int j = 0; j < 35; j++)
    for (int i = 1; i <= n; i++) {
      long long ai = (b[i] + c[i] - sum) / n;
      ssum[j] += ai & (L << j);
    }
  for (int i = 1; i <= n; i++) {
    long long ai = (b[i] + c[i] - sum) / n;
    long long x = 0;
    for (int j = 0; j < 35; j++)
      if (ai & (L << j)) x += ssum[j];
    if (x != b[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", (b[i] + c[i] - sum) / n);
  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;
long long n, b[200005], c[200005], sum = 0;
long long ssum[40];
long long L = 1;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &b[i]);
    sum += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
    sum += c[i];
  }
  if (sum % (2 * n)) {
    cout << -1;
    return 0;
  }
  sum /= 2 * n;
  for (int i = 1; i <= n; i++)
    if ((b[i] + c[i] - sum) % n || (b[i] + c[i] - sum) / n > b[i] ||
        (b[i] + c[i] - sum) / n > c[i]) {
      cout << -1;
      return 0;
    }
  long long x = 0;
  for (int i = 1; i <= n; i++)
    if (c[i] < b[i]) {
      cout << -1;
      return 0;
    }
  memset(ssum, 0, sizeof(ssum));
  for (int j = 0; j < 35; j++)
    for (int i = 1; i <= n; i++) {
      long long ai = (b[i] + c[i] - sum) / n;
      ssum[j] += ai & (L << j);
    }
  for (int i = 1; i <= n; i++) {
    long long ai = (b[i] + c[i] - sum) / n;
    long long x = 0;
    for (int j = 0; j < 35; j++)
      if (ai & (L << j)) x += ssum[j];
    if (x != b[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", (b[i] + c[i] - sum) / n);
  return 0;
}

```