#include <bits/stdc++.h>
using namespace std;
const long long size = 200002;
const long long mod = 1000000007;
const long long Inf = ~0ull >> 2;
int n, b[size], c[size], a[size];
int cnt[31];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%d", &c[i]);
  long long x = 0, y = 0;
  for (int i = 0; i < n; i++) {
    long long dt = b[i] + c[i] - b[0] - c[0];
    x += b[i] + c[i];
    if (dt % n) {
      cout << -1 << endl;
      return 0;
    }
    y += dt / n;
  }
  if ((b[0] + c[0] - y) % (2 * n)) {
    cout << -1 << endl;
    return 0;
  }
  a[0] = (b[0] + c[0] - y) / 2 / n;
  for (int i = 0; i < n; i++) {
    a[i] = a[0] + (b[i] + c[i] - b[0] - c[0]) / n;
    if (a[i] < 0) {
      cout << -1 << endl;
      return 0;
    }
    for (int j = 0; j < 31; j++) cnt[j] += ((a[i] >> j) & 1);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 31; j++) {
      if (a[i] & (1 << j)) {
        b[i] -= cnt[j] * (1 << j);
        c[i] -= n * (1 << j);
      } else
        c[i] -= cnt[j] * (1 << j);
    }
    if (b[i] || c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << ' ';
  }
}

### Prompt

In Cpp, your task is to solve 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 long long size = 200002;
const long long mod = 1000000007;
const long long Inf = ~0ull >> 2;
int n, b[size], c[size], a[size];
int cnt[31];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%d", &c[i]);
  long long x = 0, y = 0;
  for (int i = 0; i < n; i++) {
    long long dt = b[i] + c[i] - b[0] - c[0];
    x += b[i] + c[i];
    if (dt % n) {
      cout << -1 << endl;
      return 0;
    }
    y += dt / n;
  }
  if ((b[0] + c[0] - y) % (2 * n)) {
    cout << -1 << endl;
    return 0;
  }
  a[0] = (b[0] + c[0] - y) / 2 / n;
  for (int i = 0; i < n; i++) {
    a[i] = a[0] + (b[i] + c[i] - b[0] - c[0]) / n;
    if (a[i] < 0) {
      cout << -1 << endl;
      return 0;
    }
    for (int j = 0; j < 31; j++) cnt[j] += ((a[i] >> j) & 1);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 31; j++) {
      if (a[i] & (1 << j)) {
        b[i] -= cnt[j] * (1 << j);
        c[i] -= n * (1 << j);
      } else
        c[i] -= cnt[j] * (1 << j);
    }
    if (b[i] || c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << ' ';
  }
}

```