#include <bits/stdc++.h>
using namespace std;
int n;
vector<long long> b, c, a;
bool solve() {
  long long sum =
      accumulate(b.begin(), b.end(), 0ll) + accumulate(c.begin(), c.end(), 0ll);
  if (sum % (long long)(n + n) != 0) return false;
  sum /= (long long)(n + n);
  for (int i = 0; i < n; ++i) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] % n != 0) return false;
    a[i] /= n;
  }
  vector<long long> num(32, 0);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 32; ++j)
      if ((1 << j) & a[i]) ++num[j];
  }
  for (int i = 0; i < n; ++i) {
    long long bb = 0;
    long long cc = 0;
    for (int j = 0; j < 32; ++j) {
      if ((1 << j) & a[i]) {
        bb += (1ll << j) * num[j];
        cc += (1ll << j) * (long long)n;
      } else
        cc += (1ll << j) * num[j];
    }
    if (bb != b[i] || cc != c[i]) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  b.resize(n);
  for (int i = 0; i < n; ++i) cin >> b[i];
  c.resize(n);
  for (int i = 0; i < n; ++i) cin >> c[i];
  a.resize(n);
  if (solve() == false) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i) cout << a[i] << " ";
  cout << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int n;
vector<long long> b, c, a;
bool solve() {
  long long sum =
      accumulate(b.begin(), b.end(), 0ll) + accumulate(c.begin(), c.end(), 0ll);
  if (sum % (long long)(n + n) != 0) return false;
  sum /= (long long)(n + n);
  for (int i = 0; i < n; ++i) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] % n != 0) return false;
    a[i] /= n;
  }
  vector<long long> num(32, 0);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 32; ++j)
      if ((1 << j) & a[i]) ++num[j];
  }
  for (int i = 0; i < n; ++i) {
    long long bb = 0;
    long long cc = 0;
    for (int j = 0; j < 32; ++j) {
      if ((1 << j) & a[i]) {
        bb += (1ll << j) * num[j];
        cc += (1ll << j) * (long long)n;
      } else
        cc += (1ll << j) * num[j];
    }
    if (bb != b[i] || cc != c[i]) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  b.resize(n);
  for (int i = 0; i < n; ++i) cin >> b[i];
  c.resize(n);
  for (int i = 0; i < n; ++i) cin >> c[i];
  a.resize(n);
  if (solve() == false) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i) cout << a[i] << " ";
  cout << endl;
  return 0;
}

```