#include <bits/stdc++.h>
const int MAXN = (int)2e5 + 5;
long long int a[MAXN];
int b[MAXN];
int c[MAXN];
int d[MAXN];
long long int check_b[MAXN];
long long int check_c[MAXN];
int main(void) {
  int n;
  bool ok = true;
  long long int sum = 0;
  scanf(" %d", &n);
  for (int i = 0; i < n; i++) {
    scanf(" %d", &b[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf(" %d", &c[i]);
  }
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    sum += d[i];
  }
  if (sum % (2 * n) != 0) {
    ok = false;
  } else {
    sum /= (2 * n);
    for (int i = 0; i < n; i++) {
      if ((d[i] - sum) % n != 0) {
        ok = false;
      }
      a[i] = (d[i] - sum) / n;
    }
  }
  for (int bit = 30; bit >= 0; bit--) {
    int counter = 0;
    for (int i = 0; i < n; i++) {
      counter += (a[i] >> bit) & 1;
    }
    for (int i = 0; i < n; i++) {
      if ((a[i] >> bit) & 1) {
        check_b[i] += counter * (1LL << bit);
        check_c[i] += n * (1LL << bit);
      } else {
        check_c[i] += counter * (1LL << bit);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (check_b[i] != b[i] || check_c[i] != c[i]) {
      ok = false;
      break;
    }
  }
  if (ok) {
    for (int i = 0; i < n; i++) {
      if (i != 0) {
        printf(" ");
      }
      printf("%lld", a[i]);
    }
    printf("\n");
  } else {
    printf("-1\n");
  }
  return 0;
}

### Prompt

Your task is to create 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>
const int MAXN = (int)2e5 + 5;
long long int a[MAXN];
int b[MAXN];
int c[MAXN];
int d[MAXN];
long long int check_b[MAXN];
long long int check_c[MAXN];
int main(void) {
  int n;
  bool ok = true;
  long long int sum = 0;
  scanf(" %d", &n);
  for (int i = 0; i < n; i++) {
    scanf(" %d", &b[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf(" %d", &c[i]);
  }
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    sum += d[i];
  }
  if (sum % (2 * n) != 0) {
    ok = false;
  } else {
    sum /= (2 * n);
    for (int i = 0; i < n; i++) {
      if ((d[i] - sum) % n != 0) {
        ok = false;
      }
      a[i] = (d[i] - sum) / n;
    }
  }
  for (int bit = 30; bit >= 0; bit--) {
    int counter = 0;
    for (int i = 0; i < n; i++) {
      counter += (a[i] >> bit) & 1;
    }
    for (int i = 0; i < n; i++) {
      if ((a[i] >> bit) & 1) {
        check_b[i] += counter * (1LL << bit);
        check_c[i] += n * (1LL << bit);
      } else {
        check_c[i] += counter * (1LL << bit);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (check_b[i] != b[i] || check_c[i] != c[i]) {
      ok = false;
      break;
    }
  }
  if (ok) {
    for (int i = 0; i < n; i++) {
      if (i != 0) {
        printf(" ");
      }
      printf("%lld", a[i]);
    }
    printf("\n");
  } else {
    printf("-1\n");
  }
  return 0;
}

```