#include <bits/stdc++.h>
using namespace std;
const long long MX = 2e5 + 100, INF = 1e18, mod = 1e9 + 7;
long long n, m, a[MX], b[MX], c[MX], arr[MX], cnt[MX], flg, ans;
bool mark[MX];
bool palcheck(long long a) {
  for (long long i = 2; i <= sqrt(a); i++) {
    if (a % i == 0) return false;
  }
  return true;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> c[i];
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
    arr[i] = b[i] + c[i];
    sum += arr[i];
  }
  if (sum % (n * 2) == 0)
    sum /= n * 2;
  else {
    cout << -1 << endl;
    return 0;
  }
  for (long long i = 0; i < n; i++) a[i] = (arr[i] - sum) / n;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < 32; j++)
      if (a[i] & (1 << j)) cnt[j]++;
  for (long long i = 0; i < n; i++) {
    sum = 0;
    for (long long j = 0; j < 32; j++)
      if (cnt[j] && a[i] & (1 << j)) sum += (1 << j) * cnt[j];
    if (sum != c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (long long i = 0; i < n; i++) {
    sum = n * a[i];
    for (long long j = 0; j < 32; j++)
      if (!(a[i] & (1 << j)) && cnt[j]) sum += cnt[j] * (1 << j);
    if (sum != b[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  return 0;
}

### 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;
const long long MX = 2e5 + 100, INF = 1e18, mod = 1e9 + 7;
long long n, m, a[MX], b[MX], c[MX], arr[MX], cnt[MX], flg, ans;
bool mark[MX];
bool palcheck(long long a) {
  for (long long i = 2; i <= sqrt(a); i++) {
    if (a % i == 0) return false;
  }
  return true;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> c[i];
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
    arr[i] = b[i] + c[i];
    sum += arr[i];
  }
  if (sum % (n * 2) == 0)
    sum /= n * 2;
  else {
    cout << -1 << endl;
    return 0;
  }
  for (long long i = 0; i < n; i++) a[i] = (arr[i] - sum) / n;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < 32; j++)
      if (a[i] & (1 << j)) cnt[j]++;
  for (long long i = 0; i < n; i++) {
    sum = 0;
    for (long long j = 0; j < 32; j++)
      if (cnt[j] && a[i] & (1 << j)) sum += (1 << j) * cnt[j];
    if (sum != c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (long long i = 0; i < n; i++) {
    sum = n * a[i];
    for (long long j = 0; j < 32; j++)
      if (!(a[i] & (1 << j)) && cnt[j]) sum += cnt[j] * (1 << j);
    if (sum != b[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  return 0;
}

```