#include <bits/stdc++.h>
using namespace std;
int b[200005], c[200005], a[200005], bits[50];
int main() {
  int i, j, k;
  int n;
  scanf("%d", &n);
  long long s = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &b[i]);
    s += 1ll * b[i];
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &c[i]);
    s += 1ll * c[i];
  }
  if (s % (2 * n) != 0) {
    printf("-1\n");
    return 0;
  }
  s /= (2 * n);
  for (i = 0; i < n; i++) {
    if ((b[i] + c[i] - s) % n != 0) {
      printf("-1\n");
      return 0;
    }
    a[i] = (b[i] + c[i] - s) / n;
    if (a[i] < 0) {
      printf("-1\n");
      return 0;
    }
  }
  int z = 0;
  for (i = 0; i < n; i++) {
    j = a[i];
    k = 0;
    while (j != 0) {
      if (j % 2 != 0) bits[k]++;
      j /= 2;
      k++;
    }
    z = max(z, k);
  }
  for (i = 0; i < n; i++) {
    int x = 1;
    j = a[i];
    k = 0;
    while (k != z) {
      if (j % 2 == 0) {
        c[i] -= (bits[k]) * x;
      } else {
        c[i] -= (n)*x;
        b[i] -= (bits[k]) * x;
      }
      x *= 2;
      k++;
      j /= 2;
    }
  }
  for (i = 0; i < n; i++) {
    if (b[i] != 0 || c[i] != 0) {
      printf("-1\n");
      return 0;
    }
  }
  for (i = 0; i < n; i++) printf("%d ", a[i]);
  printf("\n");
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int b[200005], c[200005], a[200005], bits[50];
int main() {
  int i, j, k;
  int n;
  scanf("%d", &n);
  long long s = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &b[i]);
    s += 1ll * b[i];
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &c[i]);
    s += 1ll * c[i];
  }
  if (s % (2 * n) != 0) {
    printf("-1\n");
    return 0;
  }
  s /= (2 * n);
  for (i = 0; i < n; i++) {
    if ((b[i] + c[i] - s) % n != 0) {
      printf("-1\n");
      return 0;
    }
    a[i] = (b[i] + c[i] - s) / n;
    if (a[i] < 0) {
      printf("-1\n");
      return 0;
    }
  }
  int z = 0;
  for (i = 0; i < n; i++) {
    j = a[i];
    k = 0;
    while (j != 0) {
      if (j % 2 != 0) bits[k]++;
      j /= 2;
      k++;
    }
    z = max(z, k);
  }
  for (i = 0; i < n; i++) {
    int x = 1;
    j = a[i];
    k = 0;
    while (k != z) {
      if (j % 2 == 0) {
        c[i] -= (bits[k]) * x;
      } else {
        c[i] -= (n)*x;
        b[i] -= (bits[k]) * x;
      }
      x *= 2;
      k++;
      j /= 2;
    }
  }
  for (i = 0; i < n; i++) {
    if (b[i] != 0 || c[i] != 0) {
      printf("-1\n");
      return 0;
    }
  }
  for (i = 0; i < n; i++) printf("%d ", a[i]);
  printf("\n");
  return 0;
}

```