#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int b[maxn], c[maxn], n, cnt[100];
long long sum, a[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) cin >> c[i];
  for (int i = 1; i <= n; i++) sum += b[i];
  for (int i = 1; i <= n; i++) sum += c[i];
  if ((sum % (n * 2)) != 0) {
    cout << -1;
    return 0;
  }
  sum /= (n * 2);
  for (int i = 1; i <= n; i++) {
    long long tmp = b[i] + c[i] - sum;
    if (tmp < 0 || ((tmp % n) != 0)) {
      cout << -1;
      return 0;
    }
    a[i] = tmp / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 64; j++)
      if ((a[i] >> j) & 1LL) cnt[j]++;
  }
  for (int i = 1; i <= n; i++) {
    long long bb = 0, cc = 0;
    for (int j = 0; j < 64; j++) {
      if ((a[i] >> j) & 1LL) {
        bb += (1LL << j) * cnt[j];
        cc += 1LL * n * (1LL << j);
      } else {
        cc += 1LL * cnt[j] * (1LL << j);
      }
    }
    if (bb != 1LL * b[i] || cc != 1LL * c[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) cout << a[i] << " ";
}

### Prompt

Generate 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;
const int maxn = 2e5 + 10;
int b[maxn], c[maxn], n, cnt[100];
long long sum, a[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) cin >> c[i];
  for (int i = 1; i <= n; i++) sum += b[i];
  for (int i = 1; i <= n; i++) sum += c[i];
  if ((sum % (n * 2)) != 0) {
    cout << -1;
    return 0;
  }
  sum /= (n * 2);
  for (int i = 1; i <= n; i++) {
    long long tmp = b[i] + c[i] - sum;
    if (tmp < 0 || ((tmp % n) != 0)) {
      cout << -1;
      return 0;
    }
    a[i] = tmp / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 64; j++)
      if ((a[i] >> j) & 1LL) cnt[j]++;
  }
  for (int i = 1; i <= n; i++) {
    long long bb = 0, cc = 0;
    for (int j = 0; j < 64; j++) {
      if ((a[i] >> j) & 1LL) {
        bb += (1LL << j) * cnt[j];
        cc += 1LL * n * (1LL << j);
      } else {
        cc += 1LL * cnt[j] * (1LL << j);
      }
    }
    if (bb != 1LL * b[i] || cc != 1LL * c[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) cout << a[i] << " ";
}

```