#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
long long N, a[maxn], b[maxn], c[maxn], cnt[40];
bool check() {
  for (int i = 1; i <= N; i++) {
    long long ckb = 0, ckc = 0;
    for (int j = 0; j <= 30; j++) {
      if ((a[i] >> j) & 1) {
        ckb += cnt[j] * (1LL << j);
        ckc += (1LL << j) * N;
      } else {
        ckc += cnt[j] * (1LL << j);
      }
    }
    if (ckb != b[i] || ckc != c[i]) return false;
  }
  return true;
}
int main() {
  scanf("%lld", &N);
  long long sumbc = 0;
  for (int i = 1; i <= N; i++) {
    scanf("%lld", &b[i]);
    sumbc += b[i];
  }
  for (int i = 1; i <= N; i++) {
    scanf("%lld", &c[i]);
    sumbc += c[i];
  }
  long long suma = sumbc / 2 / N;
  for (int i = 1; i <= N; i++) {
    a[i] = (b[i] + c[i] - suma) / N;
    for (int j = 0; j <= 30; j++) {
      cnt[j] += ((a[i] >> j) & 1);
    }
  }
  if (check()) {
    for (int i = 1; i <= N; i++) printf("%lld%c", a[i], i == N ? '\n' : ' ');
  } else
    printf("-1\n");
  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 maxn = 200010;
long long N, a[maxn], b[maxn], c[maxn], cnt[40];
bool check() {
  for (int i = 1; i <= N; i++) {
    long long ckb = 0, ckc = 0;
    for (int j = 0; j <= 30; j++) {
      if ((a[i] >> j) & 1) {
        ckb += cnt[j] * (1LL << j);
        ckc += (1LL << j) * N;
      } else {
        ckc += cnt[j] * (1LL << j);
      }
    }
    if (ckb != b[i] || ckc != c[i]) return false;
  }
  return true;
}
int main() {
  scanf("%lld", &N);
  long long sumbc = 0;
  for (int i = 1; i <= N; i++) {
    scanf("%lld", &b[i]);
    sumbc += b[i];
  }
  for (int i = 1; i <= N; i++) {
    scanf("%lld", &c[i]);
    sumbc += c[i];
  }
  long long suma = sumbc / 2 / N;
  for (int i = 1; i <= N; i++) {
    a[i] = (b[i] + c[i] - suma) / N;
    for (int j = 0; j <= 30; j++) {
      cnt[j] += ((a[i] >> j) & 1);
    }
  }
  if (check()) {
    for (int i = 1; i <= N; i++) printf("%lld%c", a[i], i == N ? '\n' : ' ');
  } else
    printf("-1\n");
  return 0;
}

```