#include <bits/stdc++.h>
using namespace std;
long long int n;
long long int a[200007], b[200007], c[200007], d[200007];
long long int bb[200007], cc[200007];
long long int calc[35];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  long long int sum(0);
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    sum += d[i];
  }
  for (int i = 0; i < n; i++) {
    a[i] = (2 * n * d[i] - sum) / (2 * n * n);
    if ((2 * n * d[i] - sum) % (2 * n * n) != 0) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    long long int perm = a[i];
    long long int pos(0);
    while (perm) {
      calc[pos++] += (perm % 2);
      perm /= 2;
    }
  }
  for (int i = 0; i < n; i++) {
    long long int ci(0), bi(0);
    long long int base(1);
    for (int j = 0; j < 32; j++) {
      bi += ((a[i] & (1 << j)) != 0 ? calc[j] : 0) * base;
      ci += ((a[i] & (1 << j)) != 0 ? n : calc[j]) * base;
      base *= 2;
    }
    if (bi != b[i] || ci != c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << ' ';
  }
  cout << endl;
  return 0;
}

### Prompt

Please formulate 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;
long long int n;
long long int a[200007], b[200007], c[200007], d[200007];
long long int bb[200007], cc[200007];
long long int calc[35];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  long long int sum(0);
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    sum += d[i];
  }
  for (int i = 0; i < n; i++) {
    a[i] = (2 * n * d[i] - sum) / (2 * n * n);
    if ((2 * n * d[i] - sum) % (2 * n * n) != 0) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    long long int perm = a[i];
    long long int pos(0);
    while (perm) {
      calc[pos++] += (perm % 2);
      perm /= 2;
    }
  }
  for (int i = 0; i < n; i++) {
    long long int ci(0), bi(0);
    long long int base(1);
    for (int j = 0; j < 32; j++) {
      bi += ((a[i] & (1 << j)) != 0 ? calc[j] : 0) * base;
      ci += ((a[i] & (1 << j)) != 0 ? n : calc[j]) * base;
      base *= 2;
    }
    if (bi != b[i] || ci != c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << ' ';
  }
  cout << endl;
  return 0;
}

```