#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 2e5 + 100;
LL b[N], c[N], a[N], d[N], tot;
LL bit[62];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", b + i);
  for (int i = 1; i <= n; i++)
    scanf("%lld", d + i), c[i] = d[i] + b[i], tot += c[i];
  if (tot % (2 * n)) {
    puts("-1");
    return 0;
  }
  LL sum = tot / (2 * n);
  for (int i = 1; i <= n; i++) {
    if (c[i] < sum || (c[i] - sum) % n) {
      puts("-1");
      return 0;
    }
    a[i] = (c[i] - sum) / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 62; j++)
      if (1ll << j & a[i]) bit[j]++;
  }
  for (int i = 1; i <= n; i++) {
    int AND = a[i], OR = a[i] * n;
    for (int j = 0; j < 62; j++) {
      if (1ll << j & a[i]) {
        AND += (bit[j] - 1) * (1ll << j);
      } else {
        OR += bit[j] * (1ll << j);
      }
    }
    if (b[i] != AND || d[i] != OR) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
  puts("");
  return 0;
}

### Prompt

Your task is to create 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;
using LL = long long;
const int N = 2e5 + 100;
LL b[N], c[N], a[N], d[N], tot;
LL bit[62];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", b + i);
  for (int i = 1; i <= n; i++)
    scanf("%lld", d + i), c[i] = d[i] + b[i], tot += c[i];
  if (tot % (2 * n)) {
    puts("-1");
    return 0;
  }
  LL sum = tot / (2 * n);
  for (int i = 1; i <= n; i++) {
    if (c[i] < sum || (c[i] - sum) % n) {
      puts("-1");
      return 0;
    }
    a[i] = (c[i] - sum) / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 62; j++)
      if (1ll << j & a[i]) bit[j]++;
  }
  for (int i = 1; i <= n; i++) {
    int AND = a[i], OR = a[i] * n;
    for (int j = 0; j < 62; j++) {
      if (1ll << j & a[i]) {
        AND += (bit[j] - 1) * (1ll << j);
      } else {
        OR += bit[j] * (1ll << j);
      }
    }
    if (b[i] != AND || d[i] != OR) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
  puts("");
  return 0;
}

```