#include <bits/stdc++.h>
using namespace std;
const int nmax = 200009;
int a[nmax], b[nmax], c[nmax], sum, i, n, used[nmax], bp[nmax], cp[nmax], j,
    flag;
long long st;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &b[i]);
    st += b[i];
  }
  for (i = 1; i <= n; ++i) {
    scanf("%d", &c[i]);
    st += c[i];
  }
  if (st % (n + n) == 0)
    sum = st / (n + n);
  else {
    printf("-1\n");
    return 0;
  }
  for (i = 1; i <= n; ++i) {
    if ((b[i] + c[i] - sum) % n) {
      printf("-1\n");
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 0; j <= 30; ++j)
      if (a[i] & (1 << j)) used[j]++;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 0; j <= 30; ++j)
      if (a[i] & (1 << j)) {
        bp[i] += (1 << j) * used[j];
        cp[i] += n * (1 << j);
      } else
        cp[i] += (1 << j) * used[j];
  }
  flag = 1;
  for (i = 1; i <= n; ++i) flag &= (bp[i] == b[i] && cp[i] == c[i]);
  if (flag) {
    for (i = 1; i <= n; ++i) printf("%d\n", a[i]);
  } else
    printf("-1\n");
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 nmax = 200009;
int a[nmax], b[nmax], c[nmax], sum, i, n, used[nmax], bp[nmax], cp[nmax], j,
    flag;
long long st;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &b[i]);
    st += b[i];
  }
  for (i = 1; i <= n; ++i) {
    scanf("%d", &c[i]);
    st += c[i];
  }
  if (st % (n + n) == 0)
    sum = st / (n + n);
  else {
    printf("-1\n");
    return 0;
  }
  for (i = 1; i <= n; ++i) {
    if ((b[i] + c[i] - sum) % n) {
      printf("-1\n");
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 0; j <= 30; ++j)
      if (a[i] & (1 << j)) used[j]++;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 0; j <= 30; ++j)
      if (a[i] & (1 << j)) {
        bp[i] += (1 << j) * used[j];
        cp[i] += n * (1 << j);
      } else
        cp[i] += (1 << j) * used[j];
  }
  flag = 1;
  for (i = 1; i <= n; ++i) flag &= (bp[i] == b[i] && cp[i] == c[i]);
  if (flag) {
    for (i = 1; i <= n; ++i) printf("%d\n", a[i]);
  } else
    printf("-1\n");
  return 0;
}

```