#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const int lg = 20;
const int mod = 1e9 + 7;
long long a[N], b[N], ans[N], dp[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  if (fopen("A.inp", "r")) {
    freopen("test.inp", "r", stdin);
    freopen("test.out", "w", stdout);
  }
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> b[i];
  if (n == 1) {
    if (a[1] != b[1])
      return cout << -1, 0;
    else
      return cout << b[1], 0;
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    sum = sum + a[i] + b[i];
  }
  if (sum % (2 * n)) return cout << -1, 0;
  sum /= (2 * n);
  for (int i = 1; i <= n; i++) ans[i] = (a[i] + b[i] - sum) / n;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 30; j++) {
      if (ans[i] & (1 << j)) dp[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long tmpa = 0, tmpb = 0;
    for (int j = 0; j <= 30; j++) {
      if (ans[i] & (1 << j)) {
        tmpa += dp[j] * (1 << j);
        tmpb += n * (1 << j);
      } else
        tmpb += (dp[j] * (1 << j));
    }
    if (tmpa != a[i] || tmpb != b[i]) return cout << -1, 0;
  }
  for (int i = 1; i <= n; i++) cout << ans[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;
const int N = 1e6 + 5;
const int lg = 20;
const int mod = 1e9 + 7;
long long a[N], b[N], ans[N], dp[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  if (fopen("A.inp", "r")) {
    freopen("test.inp", "r", stdin);
    freopen("test.out", "w", stdout);
  }
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> b[i];
  if (n == 1) {
    if (a[1] != b[1])
      return cout << -1, 0;
    else
      return cout << b[1], 0;
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    sum = sum + a[i] + b[i];
  }
  if (sum % (2 * n)) return cout << -1, 0;
  sum /= (2 * n);
  for (int i = 1; i <= n; i++) ans[i] = (a[i] + b[i] - sum) / n;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 30; j++) {
      if (ans[i] & (1 << j)) dp[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long tmpa = 0, tmpb = 0;
    for (int j = 0; j <= 30; j++) {
      if (ans[i] & (1 << j)) {
        tmpa += dp[j] * (1 << j);
        tmpb += n * (1 << j);
      } else
        tmpb += (dp[j] * (1 << j));
    }
    if (tmpa != a[i] || tmpb != b[i]) return cout << -1, 0;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}

```