#include <bits/stdc++.h>
using namespace std;
int n, i, j, a[200001], b[200001], c[200001];
int mi[60];
int main() {
  scanf("%d", &n);
  long long sum = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    sum += b[i];
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    sum += c[i];
  }
  if (sum % 2) {
    printf("-1");
    return 0;
  }
  sum /= 2;
  if (sum % n) {
    printf("-1");
    return 0;
  }
  sum = sum / n;
  for (i = 1; i <= n; i++) {
    a[i] = (1ll * b[i] + c[i] - sum);
    if (a[i] % n) {
      printf("-1");
      return 0;
    } else
      a[i] /= n;
    int t = 0, t1 = a[i];
    while (a[i]) {
      mi[t] += a[i] % 2;
      a[i] /= 2;
      t++;
    }
    a[i] = t1;
  }
  for (i = 1; i <= n; i++) {
    int t1 = a[i], t = 0;
    while (t1) {
      if (t1 % 2)
        c[i] -= pow(2, t) * n;
      else
        c[i] -= pow(2, t) * mi[t];
      t++;
      t1 /= 2;
    }
    for (j = t; j <= 40; j++) c[i] -= pow(2, j) * mi[j];
    if (c[i] != 0) {
      printf("-1");
      return 0;
    }
  }
  for (i = 1; i <= n; i++) printf("%d ", a[i]);
}

### Prompt

Develop a solution in Cpp 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;
int n, i, j, a[200001], b[200001], c[200001];
int mi[60];
int main() {
  scanf("%d", &n);
  long long sum = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    sum += b[i];
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    sum += c[i];
  }
  if (sum % 2) {
    printf("-1");
    return 0;
  }
  sum /= 2;
  if (sum % n) {
    printf("-1");
    return 0;
  }
  sum = sum / n;
  for (i = 1; i <= n; i++) {
    a[i] = (1ll * b[i] + c[i] - sum);
    if (a[i] % n) {
      printf("-1");
      return 0;
    } else
      a[i] /= n;
    int t = 0, t1 = a[i];
    while (a[i]) {
      mi[t] += a[i] % 2;
      a[i] /= 2;
      t++;
    }
    a[i] = t1;
  }
  for (i = 1; i <= n; i++) {
    int t1 = a[i], t = 0;
    while (t1) {
      if (t1 % 2)
        c[i] -= pow(2, t) * n;
      else
        c[i] -= pow(2, t) * mi[t];
      t++;
      t1 /= 2;
    }
    for (j = t; j <= 40; j++) c[i] -= pow(2, j) * mi[j];
    if (c[i] != 0) {
      printf("-1");
      return 0;
    }
  }
  for (i = 1; i <= n; i++) printf("%d ", a[i]);
}

```