#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const int LN = 34;
long long n, a[N], b[N], c[N], d[N], sum, bitCount[2][LN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> 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];
    sum += d[i];
  }
  sum = sum / (2 * n);
  for (int i = 0; i < n; i++) {
    a[i] = (d[i] - sum) / n;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < LN; j++) {
      if ((1LL << j) & (a[i]))
        bitCount[1][j]++;
      else
        bitCount[0][j]++;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < LN; j++) {
      if (((1LL << j) & (a[i]))) {
        c[i] -= (bitCount[0][j] + bitCount[1][j]) * (1LL << j);
        b[i] -= (bitCount[1][j]) * (1LL << j);
      } else {
        c[i] -= (bitCount[1][j]) * (1LL << j);
      }
    }
    if (b[i] != 0 || c[i] != 0) {
      cout << "-1";
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  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;
const int N = 200005;
const int LN = 34;
long long n, a[N], b[N], c[N], d[N], sum, bitCount[2][LN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> 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];
    sum += d[i];
  }
  sum = sum / (2 * n);
  for (int i = 0; i < n; i++) {
    a[i] = (d[i] - sum) / n;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < LN; j++) {
      if ((1LL << j) & (a[i]))
        bitCount[1][j]++;
      else
        bitCount[0][j]++;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < LN; j++) {
      if (((1LL << j) & (a[i]))) {
        c[i] -= (bitCount[0][j] + bitCount[1][j]) * (1LL << j);
        b[i] -= (bitCount[1][j]) * (1LL << j);
      } else {
        c[i] -= (bitCount[1][j]) * (1LL << j);
      }
    }
    if (b[i] != 0 || c[i] != 0) {
      cout << "-1";
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  return 0;
}

```