#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000;
long long b[MAXN + 1], c[MAXN + 1];
long long a[MAXN + 1];
const int MAXBIT = 32;
int bc[MAXBIT];
bool checkBit(long long n, int bi) { return (n & (1 << bi)) != 0; }
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
  }
  long long sum = 0LL;
  for (int i = 0; i < n; ++i) {
    sum += b[i] + c[i];
  }
  if (sum % (2 * n) != 0) {
    cout << -1 << "\n";
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    if ((b[i] + c[i] - sum / (2 * n)) % n != 0) {
      cout << -1 << "\n";
      return 0;
    }
    a[i] = (b[i] + c[i] - sum / (2 * n)) / n;
  }
  for (int bi = 0; bi < MAXBIT; ++bi) {
    for (int i = 0; i < n; ++i) {
      bc[bi] += (checkBit(a[i], bi) ? 1 : 0);
    }
  }
  for (int i = 0; i < n; ++i) {
    long long constructedBi = 0LL;
    long long constructedCi = 0LL;
    for (int bi = MAXBIT - 1; bi >= 0; --bi) {
      constructedBi *= 2;
      constructedCi *= 2;
      if (checkBit(a[i], bi)) {
        constructedBi += bc[bi];
        constructedCi += n;
      } else {
        constructedCi += bc[bi];
      }
    }
    if (constructedBi != b[i] || constructedCi != c[i]) {
      cout << -1 << "\n";
      return 0;
    }
  }
  for (int i = 0; i <= n - 2; ++i) {
    cout << a[i] << " ";
  }
  cout << a[n - 1] << "\n";
  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 MAXN = 200000;
long long b[MAXN + 1], c[MAXN + 1];
long long a[MAXN + 1];
const int MAXBIT = 32;
int bc[MAXBIT];
bool checkBit(long long n, int bi) { return (n & (1 << bi)) != 0; }
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
  }
  long long sum = 0LL;
  for (int i = 0; i < n; ++i) {
    sum += b[i] + c[i];
  }
  if (sum % (2 * n) != 0) {
    cout << -1 << "\n";
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    if ((b[i] + c[i] - sum / (2 * n)) % n != 0) {
      cout << -1 << "\n";
      return 0;
    }
    a[i] = (b[i] + c[i] - sum / (2 * n)) / n;
  }
  for (int bi = 0; bi < MAXBIT; ++bi) {
    for (int i = 0; i < n; ++i) {
      bc[bi] += (checkBit(a[i], bi) ? 1 : 0);
    }
  }
  for (int i = 0; i < n; ++i) {
    long long constructedBi = 0LL;
    long long constructedCi = 0LL;
    for (int bi = MAXBIT - 1; bi >= 0; --bi) {
      constructedBi *= 2;
      constructedCi *= 2;
      if (checkBit(a[i], bi)) {
        constructedBi += bc[bi];
        constructedCi += n;
      } else {
        constructedCi += bc[bi];
      }
    }
    if (constructedBi != b[i] || constructedCi != c[i]) {
      cout << -1 << "\n";
      return 0;
    }
  }
  for (int i = 0; i <= n - 2; ++i) {
    cout << a[i] << " ";
  }
  cout << a[n - 1] << "\n";
  return 0;
}

```