#include <bits/stdc++.h>
using namespace std;
long long int a[200005], b[200005], c[200005];
int main() {
  long long int n;
  scanf("%I64d", &n);
  for (int i = 0; i < n; i++) scanf("%I64d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%I64d", &c[i]);
  long long sum = 0;
  for (int i = 0; i < n; i++) sum += b[i] + c[i];
  if (sum % (2 * n) != 0) {
    printf("-1\n");
    return 0;
  }
  sum /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = b[i] + c[i];
  for (int i = 0; i < n; i++) {
    a[i] -= sum;
    if (a[i] % n != 0) {
      printf("-1\n");
      return 0;
    }
    a[i] /= n;
  }
  int bit[100];
  memset(bit, 0, sizeof(bit));
  for (int i = 0; i < n; i++) {
    int d = a[i];
    for (int j = 0; d != 0; j++) {
      bit[j] += d & 1;
      d >>= 1;
    }
  }
  for (int i = 0; i < n; i++) {
    int d = a[i];
    sum = 0;
    for (int j = 0; d != 0; j++) {
      if (d & 1) sum += bit[j] << j;
      d /= 2;
    }
    if (sum != b[i]) {
      printf("-1\n");
      return 0;
    }
    d = a[i];
    sum = 0;
    for (int j = 0; j < 32; j++) {
      if (d & 1)
        sum += n << j;
      else
        sum += bit[j] << j;
      d /= 2;
    }
    if (sum != c[i]) {
      printf("-1\n");
      return 0;
    }
  }
  printf("%I64d", a[0]);
  for (int i = 1; i < n; i++) printf(" %I64d", a[i]);
  printf("\n");
}

### 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 int a[200005], b[200005], c[200005];
int main() {
  long long int n;
  scanf("%I64d", &n);
  for (int i = 0; i < n; i++) scanf("%I64d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%I64d", &c[i]);
  long long sum = 0;
  for (int i = 0; i < n; i++) sum += b[i] + c[i];
  if (sum % (2 * n) != 0) {
    printf("-1\n");
    return 0;
  }
  sum /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = b[i] + c[i];
  for (int i = 0; i < n; i++) {
    a[i] -= sum;
    if (a[i] % n != 0) {
      printf("-1\n");
      return 0;
    }
    a[i] /= n;
  }
  int bit[100];
  memset(bit, 0, sizeof(bit));
  for (int i = 0; i < n; i++) {
    int d = a[i];
    for (int j = 0; d != 0; j++) {
      bit[j] += d & 1;
      d >>= 1;
    }
  }
  for (int i = 0; i < n; i++) {
    int d = a[i];
    sum = 0;
    for (int j = 0; d != 0; j++) {
      if (d & 1) sum += bit[j] << j;
      d /= 2;
    }
    if (sum != b[i]) {
      printf("-1\n");
      return 0;
    }
    d = a[i];
    sum = 0;
    for (int j = 0; j < 32; j++) {
      if (d & 1)
        sum += n << j;
      else
        sum += bit[j] << j;
      d /= 2;
    }
    if (sum != c[i]) {
      printf("-1\n");
      return 0;
    }
  }
  printf("%I64d", a[0]);
  for (int i = 1; i < n; i++) printf(" %I64d", a[i]);
  printf("\n");
}

```