#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<long long> b(n), c(n);
  long long sum = 0;
  for (long long i = 0; i < n; i++) cin >> b[i], sum += b[i];
  for (long long i = 0; i < n; i++) cin >> c[i], sum += c[i];
  if (sum % (2 * n)) {
    cout << -1;
    return;
  }
  long long second = sum / (2 * n);
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) a[i] = (b[i] + c[i] - second) / n;
  for (long long i = 0; i < n; i++)
    if (b[i] + c[i] != n * a[i] + second) {
      cout << -1;
      exit(0);
    }
  vector<long long> cnt(31);
  for (long long i = 0; i < 31; i++) {
    for (long long j = 0; j < n; j++) {
      cnt[i] += ((a[j] >> i) & 1);
    }
  }
  for (long long j = 0; j < n; j++) {
    for (long long i = 0; i < 31; i++) {
      if (a[j] & (1 << i)) {
        c[j] -= (1LL << i) * n;
        b[j] -= (1LL << i) * cnt[i];
      } else {
        c[j] -= (1 << i) * cnt[i];
      }
    }
  }
  if (all_of((b).begin(), (b).end(), [](long long p) { return p == 0; }) and
      all_of((c).begin(), (c).end(), [](long long p) { return p == 0; })) {
    for (auto i : a) cout << i << " ";
  } else {
    cout << -1;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
void solve() {
  long long n;
  cin >> n;
  vector<long long> b(n), c(n);
  long long sum = 0;
  for (long long i = 0; i < n; i++) cin >> b[i], sum += b[i];
  for (long long i = 0; i < n; i++) cin >> c[i], sum += c[i];
  if (sum % (2 * n)) {
    cout << -1;
    return;
  }
  long long second = sum / (2 * n);
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) a[i] = (b[i] + c[i] - second) / n;
  for (long long i = 0; i < n; i++)
    if (b[i] + c[i] != n * a[i] + second) {
      cout << -1;
      exit(0);
    }
  vector<long long> cnt(31);
  for (long long i = 0; i < 31; i++) {
    for (long long j = 0; j < n; j++) {
      cnt[i] += ((a[j] >> i) & 1);
    }
  }
  for (long long j = 0; j < n; j++) {
    for (long long i = 0; i < 31; i++) {
      if (a[j] & (1 << i)) {
        c[j] -= (1LL << i) * n;
        b[j] -= (1LL << i) * cnt[i];
      } else {
        c[j] -= (1 << i) * cnt[i];
      }
    }
  }
  if (all_of((b).begin(), (b).end(), [](long long p) { return p == 0; }) and
      all_of((c).begin(), (c).end(), [](long long p) { return p == 0; })) {
    for (auto i : a) cout << i << " ";
  } else {
    cout << -1;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```