#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int a[maxn], b[maxn], c[maxn], p[60];
bool solve(int n) {
  long long sum = 0;
  for (int i = 1; i <= n; ++i) sum += b[i] + c[i];
  if (sum % (2 * n) != 0) return false;
  sum /= 2 * n;
  memset(p, 0, sizeof(p));
  for (int i = 1; i <= n; ++i) {
    if ((b[i] + c[i] - sum) % n != 0) return false;
    a[i] = (b[i] + c[i] - sum) / n;
    for (int j = 0; j <= 30; ++j)
      if (a[i] >> j & 1) ++p[j];
  }
  for (int i = 1; i <= n; ++i) {
    int t1 = 0, t2 = 0;
    for (int j = 0; j <= 30; ++j) {
      if (a[i] >> j & 1) {
        t1 += p[j] * (1 << j);
        t2 += n * (1 << j);
      } else {
        t2 += p[j] * (1 << j);
      }
    }
    if (t1 != b[i] || t2 != c[i]) return false;
  }
  return true;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", b + i);
  for (int i = 1; i <= n; ++i) scanf("%d", c + i);
  if (solve(n))
    for (int i = 1; i <= n; ++i) printf("%d%c", a[i], " \n"[i == n]);
  else
    puts("-1");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 maxn = 2e5 + 5;
int a[maxn], b[maxn], c[maxn], p[60];
bool solve(int n) {
  long long sum = 0;
  for (int i = 1; i <= n; ++i) sum += b[i] + c[i];
  if (sum % (2 * n) != 0) return false;
  sum /= 2 * n;
  memset(p, 0, sizeof(p));
  for (int i = 1; i <= n; ++i) {
    if ((b[i] + c[i] - sum) % n != 0) return false;
    a[i] = (b[i] + c[i] - sum) / n;
    for (int j = 0; j <= 30; ++j)
      if (a[i] >> j & 1) ++p[j];
  }
  for (int i = 1; i <= n; ++i) {
    int t1 = 0, t2 = 0;
    for (int j = 0; j <= 30; ++j) {
      if (a[i] >> j & 1) {
        t1 += p[j] * (1 << j);
        t2 += n * (1 << j);
      } else {
        t2 += p[j] * (1 << j);
      }
    }
    if (t1 != b[i] || t2 != c[i]) return false;
  }
  return true;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", b + i);
  for (int i = 1; i <= n; ++i) scanf("%d", c + i);
  if (solve(n))
    for (int i = 1; i <= n; ++i) printf("%d%c", a[i], " \n"[i == n]);
  else
    puts("-1");
  return 0;
}

```