#include <bits/stdc++.h>
using namespace std;
class Solution {
  int n;
  vector<long long> a, b, c;
  int cnt[30] = {0};

 public:
  void run() {
    cin >> n;
    a.resize(n), b.resize(n), c.resize(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    for (int i = 0; i < n; ++i) cin >> b[i];
    long long ALL = accumulate(a.begin(), a.end(), 0ll) +
                    accumulate(b.begin(), b.end(), 0ll);
    if (ALL % (2 * n)) cout << -1 << '\n', exit(0);
    long long S = ALL / (2 * n);
    for (int i = 0; i < n; ++i) {
      c[i] = (a[i] + b[i]) - S;
      if (c[i] % n) cout << -1 << '\n', exit(0);
      c[i] /= n;
      for (int j = 0; j < 30; ++j) cnt[j] += (c[i] >> j) & 1;
    }
    for (int i = 0; i < n; ++i) {
      long long cura = 0, curb = 0;
      for (int j = 0; j < 30; ++j) {
        cura += (1ll << j) * ((c[i] >> j) & 1 ? cnt[j] : 0);
        curb += (1ll << j) * ((c[i] >> j) & 1 ? n : cnt[j]);
      }
      if (cura != a[i] or curb != b[i]) cout << -1 << '\n', exit(0);
    }
    for (int i = 0; i < n; ++i) cout << c[i] << " \n"[i == n - 1];
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  Solution().run();
}

### 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;
class Solution {
  int n;
  vector<long long> a, b, c;
  int cnt[30] = {0};

 public:
  void run() {
    cin >> n;
    a.resize(n), b.resize(n), c.resize(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    for (int i = 0; i < n; ++i) cin >> b[i];
    long long ALL = accumulate(a.begin(), a.end(), 0ll) +
                    accumulate(b.begin(), b.end(), 0ll);
    if (ALL % (2 * n)) cout << -1 << '\n', exit(0);
    long long S = ALL / (2 * n);
    for (int i = 0; i < n; ++i) {
      c[i] = (a[i] + b[i]) - S;
      if (c[i] % n) cout << -1 << '\n', exit(0);
      c[i] /= n;
      for (int j = 0; j < 30; ++j) cnt[j] += (c[i] >> j) & 1;
    }
    for (int i = 0; i < n; ++i) {
      long long cura = 0, curb = 0;
      for (int j = 0; j < 30; ++j) {
        cura += (1ll << j) * ((c[i] >> j) & 1 ? cnt[j] : 0);
        curb += (1ll << j) * ((c[i] >> j) & 1 ? n : cnt[j]);
      }
      if (cura != a[i] or curb != b[i]) cout << -1 << '\n', exit(0);
    }
    for (int i = 0; i < n; ++i) cout << c[i] << " \n"[i == n - 1];
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  Solution().run();
}

```