#include <bits/stdc++.h>
using namespace std;
long long a[200010], b[200010], c[200010], d[200010];
long long bit[200010];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    memset(bit, 0, sizeof(bit));
    long long sum = 0;
    bool f = true;
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &b[i]);
      sum += b[i];
    }
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &c[i]);
      sum += c[i];
      d[i] = c[i] + b[i];
    }
    for (int i = 1; i <= n; i++) {
      a[i] = (d[i] - sum / 2 / n) / n;
      if (a[i] < 0) f = false;
    }
    if (f == false) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= 31; j++) {
        bit[j] += (a[i] & (1 << j)) ? 1 : 0;
      }
    }
    for (int i = 1; i <= n; i++) {
      long long tb = 0, tc = 0;
      for (int j = 0; j <= 31; j++) {
        if (a[i] & (1 << j)) {
          tb += bit[j] * (1 << j);
          tc += (long long)n * (1 << j);
        } else
          tc += bit[j] * (1 << j);
      }
      if (tc != c[i] || tb != b[i]) f = false;
    }
    if (f == false) {
      printf("-1\n");
      continue;
    } else
      for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
long long a[200010], b[200010], c[200010], d[200010];
long long bit[200010];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    memset(bit, 0, sizeof(bit));
    long long sum = 0;
    bool f = true;
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &b[i]);
      sum += b[i];
    }
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &c[i]);
      sum += c[i];
      d[i] = c[i] + b[i];
    }
    for (int i = 1; i <= n; i++) {
      a[i] = (d[i] - sum / 2 / n) / n;
      if (a[i] < 0) f = false;
    }
    if (f == false) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= 31; j++) {
        bit[j] += (a[i] & (1 << j)) ? 1 : 0;
      }
    }
    for (int i = 1; i <= n; i++) {
      long long tb = 0, tc = 0;
      for (int j = 0; j <= 31; j++) {
        if (a[i] & (1 << j)) {
          tb += bit[j] * (1 << j);
          tc += (long long)n * (1 << j);
        } else
          tc += bit[j] * (1 << j);
      }
      if (tc != c[i] || tb != b[i]) f = false;
    }
    if (f == false) {
      printf("-1\n");
      continue;
    } else
      for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
    printf("\n");
  }
  return 0;
}

```