#include <bits/stdc++.h>
using namespace std;
long long n, a[202020], b[202020], c[202020];
void init() {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &b[i]);
  for (int i = 0; i < n; i++) scanf("%lld", &c[i]);
}
long long sum, cnt[30];
void solve() {
  for (int i = 0; i < n; i++) sum += b[i] + c[i];
  if (sum % (n + n)) {
    puts("-1");
    exit(0);
  }
  sum /= (n + n);
  for (int i = 0; i < n; i++) {
    long long ret = b[i] + c[i];
    ret -= sum;
    if (ret < 0 || ret % n) {
      puts("-1");
      exit(0);
    }
    a[i] = ret / n;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 30; j++)
      if ((a[i] >> j) & 1ll) cnt[j]++;
  for (int i = 0; i < n; i++) {
    long long nb = 0, nc = 0;
    for (int j = 0; j < 30; j++)
      if ((a[i] >> j) & 1ll) {
        nb += (1ll << j) * cnt[j];
        nc += (1ll << j) * n;
      } else
        nc += (1ll << j) * cnt[j];
    if (nb != b[i] || nc != c[i]) {
      puts("-1");
      exit(0);
    }
  }
  for (int i = 0; i < n; i++) printf("%lld%c", a[i], " \n"[i + 1 == n]);
}
int main() {
  init();
  solve();
}

### Prompt

Your challenge is to write 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;
long long n, a[202020], b[202020], c[202020];
void init() {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &b[i]);
  for (int i = 0; i < n; i++) scanf("%lld", &c[i]);
}
long long sum, cnt[30];
void solve() {
  for (int i = 0; i < n; i++) sum += b[i] + c[i];
  if (sum % (n + n)) {
    puts("-1");
    exit(0);
  }
  sum /= (n + n);
  for (int i = 0; i < n; i++) {
    long long ret = b[i] + c[i];
    ret -= sum;
    if (ret < 0 || ret % n) {
      puts("-1");
      exit(0);
    }
    a[i] = ret / n;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 30; j++)
      if ((a[i] >> j) & 1ll) cnt[j]++;
  for (int i = 0; i < n; i++) {
    long long nb = 0, nc = 0;
    for (int j = 0; j < 30; j++)
      if ((a[i] >> j) & 1ll) {
        nb += (1ll << j) * cnt[j];
        nc += (1ll << j) * n;
      } else
        nc += (1ll << j) * cnt[j];
    if (nb != b[i] || nc != c[i]) {
      puts("-1");
      exit(0);
    }
  }
  for (int i = 0; i < n; i++) printf("%lld%c", a[i], " \n"[i + 1 == n]);
}
int main() {
  init();
  solve();
}

```