#include <bits/stdc++.h>
using namespace std;
const int N = 200500;
int n;
long long sum, a[N], b[N], c[N], cnt[61];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]), sum += b[i];
  for (int i = 1; i <= n; i++) scanf("%I64d", &c[i]), sum += c[i];
  if (sum % (2ll * n)) {
    puts("-1");
    return 0;
  }
  sum /= (2ll * n);
  for (int i = 1; i <= n; i++) {
    if (b[i] + c[i] < sum || (b[i] + c[i] - sum) % (n * 1ll)) {
      puts("-1");
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / (n * 1ll);
    for (int j = 0; j < 61; j++) {
      if (a[i] & (1ll << j)) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long bb = 0ll, cc = 0ll;
    for (int j = 0; j < 61; j++) {
      if (a[i] & (1ll << j)) {
        bb += cnt[j] * 1ll * (1ll << j);
        cc += n * 1ll * (1ll << j);
      } else
        cc += cnt[j] * 1ll * (1ll << j);
    }
    if (bb != b[i] || cc != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}

### Prompt

Please formulate 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;
const int N = 200500;
int n;
long long sum, a[N], b[N], c[N], cnt[61];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]), sum += b[i];
  for (int i = 1; i <= n; i++) scanf("%I64d", &c[i]), sum += c[i];
  if (sum % (2ll * n)) {
    puts("-1");
    return 0;
  }
  sum /= (2ll * n);
  for (int i = 1; i <= n; i++) {
    if (b[i] + c[i] < sum || (b[i] + c[i] - sum) % (n * 1ll)) {
      puts("-1");
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / (n * 1ll);
    for (int j = 0; j < 61; j++) {
      if (a[i] & (1ll << j)) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long bb = 0ll, cc = 0ll;
    for (int j = 0; j < 61; j++) {
      if (a[i] & (1ll << j)) {
        bb += cnt[j] * 1ll * (1ll << j);
        cc += n * 1ll * (1ll << j);
      } else
        cc += cnt[j] * 1ll * (1ll << j);
    }
    if (bb != b[i] || cc != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}

```