#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int b[maxn];
int c[maxn];
long long a[maxn];
long long x[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
  }
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  long long s = 0;
  for (int i = 1; i <= n; i++) {
    s += b[i] + c[i];
  }
  if (s % (2 * n) != 0)
    printf("-1\n");
  else {
    s = s / (2 * n);
    for (int i = 1; i <= n; i++) {
      a[i] = b[i] + c[i] - s;
      if (a[i] < 0 || a[i] % n != 0) {
        printf("-1\n");
        return 0;
      }
      a[i] = a[i] / n;
    }
    for (int i = 0; i <= 31; i++) {
      int all = 0;
      for (int j = 1; j <= n; j++) {
        if ((a[j] >> i) & 1) {
          all++;
        }
      }
      for (int j = 1; j <= n; j++) {
        if ((a[j] >> i) & 1) {
          x[j] += all * ((long long)1 << i);
        }
      }
    }
    for (int j = 1; j <= n; j++) {
      if (x[j] != b[j]) {
        printf("-1\n");
        return 0;
      }
    }
    for (int j = 1; j <= n; j++) printf("%I64d ", a[j]);
  }
}

### Prompt

Please formulate 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;
const int maxn = 200005;
int b[maxn];
int c[maxn];
long long a[maxn];
long long x[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
  }
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  long long s = 0;
  for (int i = 1; i <= n; i++) {
    s += b[i] + c[i];
  }
  if (s % (2 * n) != 0)
    printf("-1\n");
  else {
    s = s / (2 * n);
    for (int i = 1; i <= n; i++) {
      a[i] = b[i] + c[i] - s;
      if (a[i] < 0 || a[i] % n != 0) {
        printf("-1\n");
        return 0;
      }
      a[i] = a[i] / n;
    }
    for (int i = 0; i <= 31; i++) {
      int all = 0;
      for (int j = 1; j <= n; j++) {
        if ((a[j] >> i) & 1) {
          all++;
        }
      }
      for (int j = 1; j <= n; j++) {
        if ((a[j] >> i) & 1) {
          x[j] += all * ((long long)1 << i);
        }
      }
    }
    for (int j = 1; j <= n; j++) {
      if (x[j] != b[j]) {
        printf("-1\n");
        return 0;
      }
    }
    for (int j = 1; j <= n; j++) printf("%I64d ", a[j]);
  }
}

```