#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  long long s = 0, ss = 0, a[n], b[n], c[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) ss += b[i];
  for (int i = 0; i < n; ++i) ss += c[i];
  if (ss % (2 * n)) return cout << -1, 0;
  s = ss / (2 * n);
  int cnt[31] = {};
  for (int i = 0; i < n; ++i) {
    long long j = b[i] + c[i] - s;
    if (j % n) return cout << -1, 0;
    a[i] = j /= n;
    for (int ii = 0; j; ++ii, j >>= 1)
      if (j & 1) ++cnt[ii];
  }
  for (int i = 0; i < n; ++i) {
    long long k = 0;
    for (int j = 0; j < 31; ++j) k += (a[i] >> j & 1) * cnt[j] << j;
    if (b[i] != k) return cout << -1, 0;
    k = 0;
    for (int j = 0; j < 31; ++j) k += (a[i] >> j & 1 ? n : cnt[j]) << j;
    if (c[i] != k) return cout << -1, 0;
  }
  for (int i : a) cout << i << ' ';
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 main() {
  int n;
  cin >> n;
  long long s = 0, ss = 0, a[n], b[n], c[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) ss += b[i];
  for (int i = 0; i < n; ++i) ss += c[i];
  if (ss % (2 * n)) return cout << -1, 0;
  s = ss / (2 * n);
  int cnt[31] = {};
  for (int i = 0; i < n; ++i) {
    long long j = b[i] + c[i] - s;
    if (j % n) return cout << -1, 0;
    a[i] = j /= n;
    for (int ii = 0; j; ++ii, j >>= 1)
      if (j & 1) ++cnt[ii];
  }
  for (int i = 0; i < n; ++i) {
    long long k = 0;
    for (int j = 0; j < 31; ++j) k += (a[i] >> j & 1) * cnt[j] << j;
    if (b[i] != k) return cout << -1, 0;
    k = 0;
    for (int j = 0; j < 31; ++j) k += (a[i] >> j & 1 ? n : cnt[j]) << j;
    if (c[i] != k) return cout << -1, 0;
  }
  for (int i : a) cout << i << ' ';
}

```