#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  ;
  int n;
  cin >> n;
  vector<long long> b(n), c(n), d(n), a(n), b_con(n, 0), c_con(n, 0);
  long long sum = 0;
  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];
    sum += d[i];
  }
  sum /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = (d[i] - sum) / n;
  for (int i = 0; i < n; i++) {
    if (a[i] < 0) {
      cout << -1 << endl;
      return 0;
    }
  }
  vector<vector<int> > bits(n, vector<int>(31, 0));
  vector<int> count(31, 0);
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      if ((a[j] & ((long long)1 << i)) != 0) {
        bits[j][i] = 1;
      }
    }
  }
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      if (bits[j][i] == 1) {
        count[i] += 1;
      }
    }
  }
  for (int j = 0; j < n; j++) {
    for (int i = 0; i < 31; i++) {
      if (bits[j][i] == 0) {
        c_con[j] += count[i] << (long long)i;
      } else if (bits[j][i] == 1) {
        b_con[j] += count[i] << (long long)i;
        c_con[j] += n << (long long)i;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (b[i] != b_con[i] or c[i] != c_con[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}

### Prompt

Generate a CPP solution 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 main() {
  ios_base::sync_with_stdio(0);
  ;
  int n;
  cin >> n;
  vector<long long> b(n), c(n), d(n), a(n), b_con(n, 0), c_con(n, 0);
  long long sum = 0;
  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];
    sum += d[i];
  }
  sum /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = (d[i] - sum) / n;
  for (int i = 0; i < n; i++) {
    if (a[i] < 0) {
      cout << -1 << endl;
      return 0;
    }
  }
  vector<vector<int> > bits(n, vector<int>(31, 0));
  vector<int> count(31, 0);
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      if ((a[j] & ((long long)1 << i)) != 0) {
        bits[j][i] = 1;
      }
    }
  }
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      if (bits[j][i] == 1) {
        count[i] += 1;
      }
    }
  }
  for (int j = 0; j < n; j++) {
    for (int i = 0; i < 31; i++) {
      if (bits[j][i] == 0) {
        c_con[j] += count[i] << (long long)i;
      } else if (bits[j][i] == 1) {
        b_con[j] += count[i] << (long long)i;
        c_con[j] += n << (long long)i;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (b[i] != b_con[i] or c[i] != c_con[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}

```