#include <bits/stdc++.h>
using namespace std;
int N;
long long a[210000], b[210000], c[210000], d[210000], B[210000], C[210000];
long long no[40];
long long sum;
int32_t main() {
  ios_base::sync_with_stdio(false);
  if (fopen("cf734f.in", "r")) {
    freopen("cf734f.in", "r", stdin);
    freopen("cf734f.out", "w", stdout);
  }
  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];
  }
  for (int i = 0; i < N; i++) {
    sum += d[i];
  }
  if (sum % (2 * N)) {
    cout << "-1\n";
    return 0;
  }
  sum /= (2 * N);
  for (int i = 0; i < N; i++) {
    if ((d[i] - sum) % N) {
      cout << "-1\n";
      return 0;
    }
    a[i] = (d[i] - sum) / N;
  }
  for (int i = 0; i < N; i++) {
    long long val = a[i];
    for (int j = 0; j < 32; j++) {
      if (val & (1ll << j)) {
        no[j]++;
      }
    }
  }
  for (int i = 0; i < N; i++) {
    long long val = a[i];
    for (int j = 0; j < 32; j++) {
      if (val & (1ll << j)) {
        B[i] += ((1ll << j) * (no[j]));
        C[i] += ((1ll << j) * (N));
      } else {
        C[i] += ((1ll << j) * (no[j]));
      }
    }
  }
  for (int i = 0; i < N; i++) {
    if (B[i] != b[i] || C[i] != c[i]) {
      cout << "-1\n";
      return 0;
    }
  }
  cout << a[0];
  for (int i = 1; i < N; i++) {
    cout << ' ' << a[i];
  }
  cout << '\n';
  return 0;
}

### 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;
int N;
long long a[210000], b[210000], c[210000], d[210000], B[210000], C[210000];
long long no[40];
long long sum;
int32_t main() {
  ios_base::sync_with_stdio(false);
  if (fopen("cf734f.in", "r")) {
    freopen("cf734f.in", "r", stdin);
    freopen("cf734f.out", "w", stdout);
  }
  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];
  }
  for (int i = 0; i < N; i++) {
    sum += d[i];
  }
  if (sum % (2 * N)) {
    cout << "-1\n";
    return 0;
  }
  sum /= (2 * N);
  for (int i = 0; i < N; i++) {
    if ((d[i] - sum) % N) {
      cout << "-1\n";
      return 0;
    }
    a[i] = (d[i] - sum) / N;
  }
  for (int i = 0; i < N; i++) {
    long long val = a[i];
    for (int j = 0; j < 32; j++) {
      if (val & (1ll << j)) {
        no[j]++;
      }
    }
  }
  for (int i = 0; i < N; i++) {
    long long val = a[i];
    for (int j = 0; j < 32; j++) {
      if (val & (1ll << j)) {
        B[i] += ((1ll << j) * (no[j]));
        C[i] += ((1ll << j) * (N));
      } else {
        C[i] += ((1ll << j) * (no[j]));
      }
    }
  }
  for (int i = 0; i < N; i++) {
    if (B[i] != b[i] || C[i] != c[i]) {
      cout << "-1\n";
      return 0;
    }
  }
  cout << a[0];
  for (int i = 1; i < N; i++) {
    cout << ' ' << a[i];
  }
  cout << '\n';
  return 0;
}

```