#include <bits/stdc++.h>
using namespace std;
const int Nero = 0;
const int N = 200000 + 100;
int a[N], b[N], c[N], bb[N], cc[N];
int n;
int bit[N][35];
int bitsum[35];
int main() {
  long long sum = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", b + i);
    sum += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", c + i);
    sum += c[i];
  }
  if (sum % (2 * n)) {
    puts("-1");
    return Nero;
  }
  sum /= (2 * n);
  for (int i = 1; i <= n; i++) {
    long long temp = (long long)(b[i] + c[i]) - sum;
    if (temp < 0 || temp % n) {
      puts("-1");
      return Nero;
    } else
      a[i] = temp / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 31; j++) {
      bit[i][j] = (a[i] & (1 << j)) ? 1 : 0;
      bitsum[j] += bit[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 31; j++) {
      bb[i] += (bit[i][j] ? bitsum[j] : 0) << j;
      cc[i] += (bit[i][j] ? n : bitsum[j]) << j;
    }
  }
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    if (b[i] != bb[i] || c[i] != cc[i]) {
      flag = 0;
      break;
    }
  }
  if (flag == 0)
    printf("-1\n");
  else
    for (int i = 1; i <= n; i++) printf("%d%c", a[i], i == n ? '\n' : ' ');
  return Nero;
}

### Prompt

Please create a solution in cpp 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;
const int Nero = 0;
const int N = 200000 + 100;
int a[N], b[N], c[N], bb[N], cc[N];
int n;
int bit[N][35];
int bitsum[35];
int main() {
  long long sum = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", b + i);
    sum += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", c + i);
    sum += c[i];
  }
  if (sum % (2 * n)) {
    puts("-1");
    return Nero;
  }
  sum /= (2 * n);
  for (int i = 1; i <= n; i++) {
    long long temp = (long long)(b[i] + c[i]) - sum;
    if (temp < 0 || temp % n) {
      puts("-1");
      return Nero;
    } else
      a[i] = temp / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 31; j++) {
      bit[i][j] = (a[i] & (1 << j)) ? 1 : 0;
      bitsum[j] += bit[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 31; j++) {
      bb[i] += (bit[i][j] ? bitsum[j] : 0) << j;
      cc[i] += (bit[i][j] ? n : bitsum[j]) << j;
    }
  }
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    if (b[i] != bb[i] || c[i] != cc[i]) {
      flag = 0;
      break;
    }
  }
  if (flag == 0)
    printf("-1\n");
  else
    for (int i = 1; i <= n; i++) printf("%d%c", a[i], i == n ? '\n' : ' ');
  return Nero;
}

```