#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200010;
long long A[MAXN], B[MAXN], C[MAXN];
long long D[31];
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 = 0;
  for (int i = 0; i < N; ++i) sum += B[i] + C[i];
  if (sum % (N + N) != 0) {
    cout << -1 << endl;
    return 0;
  }
  sum /= N + N;
  for (int i = 0; i < N; ++i) {
    A[i] = B[i] + C[i] - sum;
    if (A[i] % N != 0) {
      cout << -1 << endl;
      return 0;
    }
    A[i] /= N;
    for (int j = 0; j < 31; ++j) D[j] += ((A[i] & (1 << j)) > 0);
  }
  for (int i = 0; i < N; ++i) {
    long long BB = 0, CC = 0;
    for (int j = 0; j < 31; ++j) {
      if (A[i] & (1 << j)) {
        BB += D[j] * (1LL << j);
        CC += N * (1LL << j);
      } else {
        CC += D[j] * (1LL << j);
      }
    }
    if (BB != B[i] || CC != C[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < N; ++i) cout << A[i] << " ";
  cout << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 = 200010;
long long A[MAXN], B[MAXN], C[MAXN];
long long D[31];
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 = 0;
  for (int i = 0; i < N; ++i) sum += B[i] + C[i];
  if (sum % (N + N) != 0) {
    cout << -1 << endl;
    return 0;
  }
  sum /= N + N;
  for (int i = 0; i < N; ++i) {
    A[i] = B[i] + C[i] - sum;
    if (A[i] % N != 0) {
      cout << -1 << endl;
      return 0;
    }
    A[i] /= N;
    for (int j = 0; j < 31; ++j) D[j] += ((A[i] & (1 << j)) > 0);
  }
  for (int i = 0; i < N; ++i) {
    long long BB = 0, CC = 0;
    for (int j = 0; j < 31; ++j) {
      if (A[i] & (1 << j)) {
        BB += D[j] * (1LL << j);
        CC += N * (1LL << j);
      } else {
        CC += D[j] * (1LL << j);
      }
    }
    if (BB != B[i] || CC != C[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < N; ++i) cout << A[i] << " ";
  cout << endl;
  return 0;
}

```