#include <bits/stdc++.h>
using namespace std;
long long sum, now;
int a[200005], b[200005], c[200005], num[35];
int i, j, k, n;
bool vis[200005][35];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (i = 1; i <= n; i++) sum += b[i] + c[i];
  sum /= 2 * n;
  for (i = 1; i <= n; i++) a[i] = (b[i] + c[i] - sum) / n;
  for (i = 1; i <= n; i++)
    for (j = 0; (1 << j) <= a[i]; j++)
      if (a[i] & (1 << j)) num[j]++, vis[i][j] = 1;
  for (i = 1; i <= n; i++) {
    now = 0;
    for (j = 0; (1 << j) <= a[i]; j++)
      if (vis[i][j]) now += 1ll * (1 << j) * num[j];
    if (now != b[i]) {
      puts("-1");
      return 0;
    }
    now = 0;
    for (j = 0; j <= 30; j++)
      if (vis[i][j])
        now += 1ll * (1 << j) * n;
      else
        now += 1ll * (1 << j) * num[j];
    if (now != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (i = 1; i <= n; i++) printf("%d ", a[i]);
  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;
long long sum, now;
int a[200005], b[200005], c[200005], num[35];
int i, j, k, n;
bool vis[200005][35];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (i = 1; i <= n; i++) sum += b[i] + c[i];
  sum /= 2 * n;
  for (i = 1; i <= n; i++) a[i] = (b[i] + c[i] - sum) / n;
  for (i = 1; i <= n; i++)
    for (j = 0; (1 << j) <= a[i]; j++)
      if (a[i] & (1 << j)) num[j]++, vis[i][j] = 1;
  for (i = 1; i <= n; i++) {
    now = 0;
    for (j = 0; (1 << j) <= a[i]; j++)
      if (vis[i][j]) now += 1ll * (1 << j) * num[j];
    if (now != b[i]) {
      puts("-1");
      return 0;
    }
    now = 0;
    for (j = 0; j <= 30; j++)
      if (vis[i][j])
        now += 1ll * (1 << j) * n;
      else
        now += 1ll * (1 << j) * num[j];
    if (now != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}

```