#include <bits/stdc++.h>
using namespace std;
const int ND = 32;
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> b(n);
  vector<int> c(n);
  vector<int> d(n);
  for (int i = 0; i < n; ++i) cin >> b[i];
  for (int i = 0; i < n; ++i) cin >> c[i];
  for (int i = 0; i < n; ++i) d[i] = b[i] + c[i];
  long long dsum = 0;
  for (int i = 0; i < n; ++i) dsum += d[i];
  long long asum = dsum / 2 / n;
  for (int i = 0; i < n; ++i) a[i] = (d[i] - (int)asum) / n;
  int num[ND];
  for (int i = 0; i < ND; ++i) {
    num[i] = 0;
    for (int j = 0; j < n; ++j) {
      if (a[j] & (1 << i)) ++num[i];
    }
  }
  vector<int> bt[ND];
  vector<int> ct[ND];
  for (int i = 0; i < ND; ++i) {
    bt[i].resize(n);
    ct[i].resize(n);
    for (int j = 0; j < n; ++j) {
      if (a[j] & (1 << i)) {
        bt[i][j] = num[i];
        ct[i][j] = n;
      } else {
        bt[i][j] = 0;
        ct[i][j] = num[i];
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int curr_b = 0;
    int curr_c = 0;
    for (int j = 0; j < ND; ++j) {
      curr_b += (bt[j][i] << j);
      curr_c += (ct[j][i] << j);
    }
    if (b[i] != curr_b || c[i] != curr_c) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; ++i) cout << a[i] << " ";
  cout << endl;
  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 int ND = 32;
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> b(n);
  vector<int> c(n);
  vector<int> d(n);
  for (int i = 0; i < n; ++i) cin >> b[i];
  for (int i = 0; i < n; ++i) cin >> c[i];
  for (int i = 0; i < n; ++i) d[i] = b[i] + c[i];
  long long dsum = 0;
  for (int i = 0; i < n; ++i) dsum += d[i];
  long long asum = dsum / 2 / n;
  for (int i = 0; i < n; ++i) a[i] = (d[i] - (int)asum) / n;
  int num[ND];
  for (int i = 0; i < ND; ++i) {
    num[i] = 0;
    for (int j = 0; j < n; ++j) {
      if (a[j] & (1 << i)) ++num[i];
    }
  }
  vector<int> bt[ND];
  vector<int> ct[ND];
  for (int i = 0; i < ND; ++i) {
    bt[i].resize(n);
    ct[i].resize(n);
    for (int j = 0; j < n; ++j) {
      if (a[j] & (1 << i)) {
        bt[i][j] = num[i];
        ct[i][j] = n;
      } else {
        bt[i][j] = 0;
        ct[i][j] = num[i];
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int curr_b = 0;
    int curr_c = 0;
    for (int j = 0; j < ND; ++j) {
      curr_b += (bt[j][i] << j);
      curr_c += (ct[j][i] << j);
    }
    if (b[i] != curr_b || c[i] != curr_c) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; ++i) cout << a[i] << " ";
  cout << endl;
  return 0;
}

```