#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<long long> b(n), c(n), a(n);
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &b[i]);
    sum += b[i];
  }
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &c[i]);
    sum += c[i];
  }
  sum /= (n * 2);
  for (int i = 0; i < n; ++i) a[i] = (b[i] + c[i] - sum) / n;
  bool ok = true;
  long long total_bits[60] = {};
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 60; ++j)
      if ((1LL << j) & a[i]) ++total_bits[j];
  for (int i = 0; i < n; ++i) {
    long long b_exp = 0, c_exp = 0;
    for (int j = 0; j < 60; ++j)
      if ((1LL << j) & a[i]) {
        b_exp += total_bits[j] << j;
        c_exp += n << j;
      } else {
        b_exp += 0;
        c_exp += total_bits[j] << j;
      }
    if (b_exp != b[i] || c_exp != c[i]) {
      ok = false;
      break;
    }
  }
  if (!ok) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i < n; ++i) printf("%lld%c", a[i], " \n"[i == n - 1]);
}

### Prompt

Create a solution in cpp for 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;
int main() {
  int n;
  cin >> n;
  vector<long long> b(n), c(n), a(n);
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &b[i]);
    sum += b[i];
  }
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &c[i]);
    sum += c[i];
  }
  sum /= (n * 2);
  for (int i = 0; i < n; ++i) a[i] = (b[i] + c[i] - sum) / n;
  bool ok = true;
  long long total_bits[60] = {};
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 60; ++j)
      if ((1LL << j) & a[i]) ++total_bits[j];
  for (int i = 0; i < n; ++i) {
    long long b_exp = 0, c_exp = 0;
    for (int j = 0; j < 60; ++j)
      if ((1LL << j) & a[i]) {
        b_exp += total_bits[j] << j;
        c_exp += n << j;
      } else {
        b_exp += 0;
        c_exp += total_bits[j] << j;
      }
    if (b_exp != b[i] || c_exp != c[i]) {
      ok = false;
      break;
    }
  }
  if (!ok) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i < n; ++i) printf("%lld%c", a[i], " \n"[i == n - 1]);
}

```