#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
long long sumb, sumc;
int n, a[N], b[N], c[N], cnt[30];
bool check() {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 30; j++) {
      if (a[i] & 1 << j) cnt[j]++;
    }
  }
  for (int i = 0; i < n; i++) {
    long long sum = 0;
    for (int j = 0; j < 30; j++) {
      if (a[i] & 1 << j) sum += cnt[j] * (1 << j);
    }
    if (sum != b[i]) return 0;
  }
  return 1;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", b + i), sumb += b[i];
  for (int i = 0; i < n; i++) scanf("%d", c + i), sumc += c[i];
  for (int i = 0; i < n; i++) {
    long long buf = b[i] + c[i] - (sumb + sumc) / 2 / n;
    if ((sumb + sumc) % (2 * n) || buf % n) {
      puts("-1");
      return 0;
    }
    a[i] = buf / n;
  }
  if (!check())
    puts("-1");
  else {
    for (int i = 0; i < n; i++) {
      printf("%d%c", a[i], " \n"[i == n - 1]);
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 = 2e5 + 7;
long long sumb, sumc;
int n, a[N], b[N], c[N], cnt[30];
bool check() {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 30; j++) {
      if (a[i] & 1 << j) cnt[j]++;
    }
  }
  for (int i = 0; i < n; i++) {
    long long sum = 0;
    for (int j = 0; j < 30; j++) {
      if (a[i] & 1 << j) sum += cnt[j] * (1 << j);
    }
    if (sum != b[i]) return 0;
  }
  return 1;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", b + i), sumb += b[i];
  for (int i = 0; i < n; i++) scanf("%d", c + i), sumc += c[i];
  for (int i = 0; i < n; i++) {
    long long buf = b[i] + c[i] - (sumb + sumc) / 2 / n;
    if ((sumb + sumc) % (2 * n) || buf % n) {
      puts("-1");
      return 0;
    }
    a[i] = buf / n;
  }
  if (!check())
    puts("-1");
  else {
    for (int i = 0; i < n; i++) {
      printf("%d%c", a[i], " \n"[i == n - 1]);
    }
  }
  return 0;
}

```