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

### Prompt

Construct a cpp code solution to the problem outlined:

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

```