#include <bits/stdc++.h>
const int sz = 200010;
long long b[sz], c[sz];
long long ans[sz];
long long an[60], oo[60];
int main() {
  long long n, k;
  long long sum = 0;
  scanf("%I64d", &n);
  for (k = 1; k <= n; k++) {
    scanf("%I64d", &b[k]);
    sum += b[k];
  }
  for (k = 1; k <= n; k++) {
    scanf("%I64d", &c[k]);
    sum += c[k];
  }
  if (sum % (2 * n) != 0) {
    puts("-1");
    return 0;
  }
  sum /= (2 * n);
  for (k = 1; k <= n; k++) {
    long long p = b[k] + c[k] - sum;
    if (p < 0 || p % n != 0) {
      puts("-1");
      return 0;
    }
    ans[k] = p / n;
  }
  for (k = 1; k <= n; k++) {
    long long tmp = ans[k];
    long long t = 0;
    while (tmp) {
      if (tmp & 1) {
        an[t]++;
      }
      t++;
      tmp /= 2;
    }
  }
  for (k = 1; k <= n; k++) {
    long long qq = 0;
    long long tmp = ans[k];
    long long t = 0;
    while (tmp) {
      if (tmp & 1) qq += (an[t] * (1 << t));
      tmp /= 2;
      t++;
    }
    if (qq != b[k]) {
      puts("-1");
      return 0;
    }
  }
  for (k = 1; k <= n; k++) {
    long long qq = 0;
    long long tmp = ans[k];
    long long t = 0;
    for (int i = 1; i <= 50; i++) {
      if (tmp & 1)
        qq += (n * (1 << t));
      else
        qq += (an[t] * (1 << t));
      tmp /= 2;
      t++;
    }
    if (qq != c[k]) {
      puts("-1");
      return 0;
    }
  }
  for (k = 1; k <= n; k++) printf("%I64d ", ans[k]);
}

### 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>
const int sz = 200010;
long long b[sz], c[sz];
long long ans[sz];
long long an[60], oo[60];
int main() {
  long long n, k;
  long long sum = 0;
  scanf("%I64d", &n);
  for (k = 1; k <= n; k++) {
    scanf("%I64d", &b[k]);
    sum += b[k];
  }
  for (k = 1; k <= n; k++) {
    scanf("%I64d", &c[k]);
    sum += c[k];
  }
  if (sum % (2 * n) != 0) {
    puts("-1");
    return 0;
  }
  sum /= (2 * n);
  for (k = 1; k <= n; k++) {
    long long p = b[k] + c[k] - sum;
    if (p < 0 || p % n != 0) {
      puts("-1");
      return 0;
    }
    ans[k] = p / n;
  }
  for (k = 1; k <= n; k++) {
    long long tmp = ans[k];
    long long t = 0;
    while (tmp) {
      if (tmp & 1) {
        an[t]++;
      }
      t++;
      tmp /= 2;
    }
  }
  for (k = 1; k <= n; k++) {
    long long qq = 0;
    long long tmp = ans[k];
    long long t = 0;
    while (tmp) {
      if (tmp & 1) qq += (an[t] * (1 << t));
      tmp /= 2;
      t++;
    }
    if (qq != b[k]) {
      puts("-1");
      return 0;
    }
  }
  for (k = 1; k <= n; k++) {
    long long qq = 0;
    long long tmp = ans[k];
    long long t = 0;
    for (int i = 1; i <= 50; i++) {
      if (tmp & 1)
        qq += (n * (1 << t));
      else
        qq += (an[t] * (1 << t));
      tmp /= 2;
      t++;
    }
    if (qq != c[k]) {
      puts("-1");
      return 0;
    }
  }
  for (k = 1; k <= n; k++) printf("%I64d ", ans[k]);
}

```