#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
long long a[maxn], b[maxn], c[maxn];
bool check(long long n, long long sum) {
  for (int i = 1; i <= n; i++) {
    if (b[i] + c[i] != n * a[i] + sum) {
      return false;
    }
  }
  for (int i = 30; i >= 0; i--) {
    int num = 0;
    for (int j = 1; j <= n; j++) {
      if (a[j] & (1 << i)) {
        num++;
      }
    }
    for (int j = 1; j <= n; j++) {
      if (a[j] & (1 << i)) {
        c[j] -= (1 << i) * n;
        b[j] -= (1 << i) * num;
      } else {
        c[j] -= (1 << i) * num;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (c[i] || b[i]) return false;
  }
  return true;
}
int main() {
  srand(time(NULL));
  long long n;
  scanf("%lld", &n);
  long long sum1 = 0, sum2 = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &b[i]);
    sum1 += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
    sum2 += c[i];
  }
  long long sum = (sum1 + sum2) / 2 / n;
  for (int i = 1; i <= n; i++) {
    a[i] = ((b[i] + c[i]) - sum) / n;
  }
  if (check(n, sum)) {
    for (int i = 1; i <= n; i++) {
      printf("%lld ", a[i]);
    }
  } 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 = 2e5 + 7;
long long a[maxn], b[maxn], c[maxn];
bool check(long long n, long long sum) {
  for (int i = 1; i <= n; i++) {
    if (b[i] + c[i] != n * a[i] + sum) {
      return false;
    }
  }
  for (int i = 30; i >= 0; i--) {
    int num = 0;
    for (int j = 1; j <= n; j++) {
      if (a[j] & (1 << i)) {
        num++;
      }
    }
    for (int j = 1; j <= n; j++) {
      if (a[j] & (1 << i)) {
        c[j] -= (1 << i) * n;
        b[j] -= (1 << i) * num;
      } else {
        c[j] -= (1 << i) * num;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (c[i] || b[i]) return false;
  }
  return true;
}
int main() {
  srand(time(NULL));
  long long n;
  scanf("%lld", &n);
  long long sum1 = 0, sum2 = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &b[i]);
    sum1 += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
    sum2 += c[i];
  }
  long long sum = (sum1 + sum2) / 2 / n;
  for (int i = 1; i <= n; i++) {
    a[i] = ((b[i] + c[i]) - sum) / n;
  }
  if (check(n, sum)) {
    for (int i = 1; i <= n; i++) {
      printf("%lld ", a[i]);
    }
  } else {
    printf("-1\n");
  }
  return 0;
}

```