#include <bits/stdc++.h>
using namespace std;
int const inf = 1000 * 1000 * 1000;
long long const inf64 = 1ll * inf * inf;
int const N = 200005;
int n;
int a[N];
int b[N];
int c[N];
int main() {
  scanf("%d", &n);
  long long tot = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    tot += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    tot += c[i];
  }
  if (tot % (2ll * n)) {
    puts("-1");
    return 0;
  }
  tot /= (2ll * n);
  for (int i = 1; i <= n; i++) {
    long long value = b[i] + c[i] - tot;
    if (value < 0 || value % n) {
      puts("-1");
      return 0;
    }
    value /= n;
    if (value < 0 || value > c[i]) {
      puts("-1");
      return 0;
    }
    a[i] = (int)value;
  }
  long long delB = 0;
  long long delC = 0;
  for (int cnt, bit = 0; bit <= 30; bit++) {
    cnt = 0;
    for (int i = 1; i <= n; i++) {
      if ((a[i] >> bit) & 1) cnt++;
    }
    for (int i = 1; i <= n; i++) {
      if ((a[i] >> bit) & 1) {
        delB = (1ll << bit) * cnt;
        delC = (1ll << bit) * n;
      } else {
        delB = 0;
        delC = (1ll << bit) * cnt;
      }
      if (delB > b[i] || delC > c[i]) {
        puts("-1");
        return 0;
      } else {
        b[i] -= delB;
        c[i] -= delC;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}

### Prompt

Generate 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;
int const inf = 1000 * 1000 * 1000;
long long const inf64 = 1ll * inf * inf;
int const N = 200005;
int n;
int a[N];
int b[N];
int c[N];
int main() {
  scanf("%d", &n);
  long long tot = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    tot += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    tot += c[i];
  }
  if (tot % (2ll * n)) {
    puts("-1");
    return 0;
  }
  tot /= (2ll * n);
  for (int i = 1; i <= n; i++) {
    long long value = b[i] + c[i] - tot;
    if (value < 0 || value % n) {
      puts("-1");
      return 0;
    }
    value /= n;
    if (value < 0 || value > c[i]) {
      puts("-1");
      return 0;
    }
    a[i] = (int)value;
  }
  long long delB = 0;
  long long delC = 0;
  for (int cnt, bit = 0; bit <= 30; bit++) {
    cnt = 0;
    for (int i = 1; i <= n; i++) {
      if ((a[i] >> bit) & 1) cnt++;
    }
    for (int i = 1; i <= n; i++) {
      if ((a[i] >> bit) & 1) {
        delB = (1ll << bit) * cnt;
        delC = (1ll << bit) * n;
      } else {
        delB = 0;
        delC = (1ll << bit) * cnt;
      }
      if (delB > b[i] || delC > c[i]) {
        puts("-1");
        return 0;
      } else {
        b[i] -= delB;
        c[i] -= delC;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}

```