#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  int* a = new int[n];
  int* b = new int[n];
  int* c = new int[n];
  int* d = new int[n];
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> c[i];
  }
  long long sum = 0;
  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;
    }
  }
  const int max_int_bits = 32;
  int bits[max_int_bits][n];
  for (int i = 0; i < max_int_bits - 1; i++) {
    for (int j = 0; j < n; j++) {
      if ((a[j] & (1LL << i)) == 0)
        bits[i][j] = 0;
      else
        bits[i][j] = 1;
    }
  }
  int* kbits = new int[max_int_bits - 1];
  memset(kbits, 0, sizeof(int) * (max_int_bits - 1));
  for (int i = 0; i < max_int_bits - 1; i++) {
    for (int j = 0; j < n; j++) {
      kbits[i] += bits[i][j];
    }
  }
  int* b1 = new int[n];
  int* c1 = new int[n];
  memset(b1, 0, sizeof(int) * n);
  memset(c1, 0, sizeof(int) * n);
  for (int i = 0; i < max_int_bits - 1; i++) {
    for (int j = 0; j < n; j++) {
      if (bits[i][j] == 0) {
        b1[j] += 0;
        c1[j] += (kbits[i] << i);
      } else {
        b1[j] += (kbits[i] << i);
        c1[j] += (n << i);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (b1[i] != b[i] || c1[i] != c[i]) {
      cout << -1 << endl;
      return;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
int main() {
  solve();
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
void solve() {
  int n;
  cin >> n;
  int* a = new int[n];
  int* b = new int[n];
  int* c = new int[n];
  int* d = new int[n];
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> c[i];
  }
  long long sum = 0;
  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;
    }
  }
  const int max_int_bits = 32;
  int bits[max_int_bits][n];
  for (int i = 0; i < max_int_bits - 1; i++) {
    for (int j = 0; j < n; j++) {
      if ((a[j] & (1LL << i)) == 0)
        bits[i][j] = 0;
      else
        bits[i][j] = 1;
    }
  }
  int* kbits = new int[max_int_bits - 1];
  memset(kbits, 0, sizeof(int) * (max_int_bits - 1));
  for (int i = 0; i < max_int_bits - 1; i++) {
    for (int j = 0; j < n; j++) {
      kbits[i] += bits[i][j];
    }
  }
  int* b1 = new int[n];
  int* c1 = new int[n];
  memset(b1, 0, sizeof(int) * n);
  memset(c1, 0, sizeof(int) * n);
  for (int i = 0; i < max_int_bits - 1; i++) {
    for (int j = 0; j < n; j++) {
      if (bits[i][j] == 0) {
        b1[j] += 0;
        c1[j] += (kbits[i] << i);
      } else {
        b1[j] += (kbits[i] << i);
        c1[j] += (n << i);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (b1[i] != b[i] || c1[i] != c[i]) {
      cout << -1 << endl;
      return;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
int main() {
  solve();
  return 0;
}

```