#include <bits/stdc++.h>
using namespace std;
long long n, i, j, a[200002], b[200002], c[200002], cnt[200002];
long long read() {
  char c = getchar();
  long long w = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c <= '9' && c >= '0') {
    w = w * 10 + c - '0';
    c = getchar();
  }
  return w;
}
signed main() {
  n = read();
  for (i = 1; i <= n; i++) b[i] = read();
  for (i = 1; i <= n; i++) c[i] = read();
  long long sum = 0;
  for (i = 1; i <= n; i++) sum += b[i] + c[i];
  sum /= 2 * n;
  for (i = 1; i <= n; i++) {
    if ((b[i] + c[i] - sum) % n != 0) {
      puts("-1");
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; (1 << j) <= a[i]; j++) {
      if (a[i] & (1 << j)) cnt[j]++;
    }
  }
  for (i = 1; i <= n; i++) {
    long long b1 = 0, c1 = sum;
    for (j = 0; (1 << j) <= a[i]; j++) {
      if (a[i] & (1 << j))
        b1 += (1 << j) * cnt[j], c1 += (1 << j) * (n - cnt[j]);
    }
    if (b1 != b[i] || c1 != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (i = 1; i <= n; i++) printf("%lld ", a[i]);
  puts("");
  return 0;
}

### 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;
long long n, i, j, a[200002], b[200002], c[200002], cnt[200002];
long long read() {
  char c = getchar();
  long long w = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c <= '9' && c >= '0') {
    w = w * 10 + c - '0';
    c = getchar();
  }
  return w;
}
signed main() {
  n = read();
  for (i = 1; i <= n; i++) b[i] = read();
  for (i = 1; i <= n; i++) c[i] = read();
  long long sum = 0;
  for (i = 1; i <= n; i++) sum += b[i] + c[i];
  sum /= 2 * n;
  for (i = 1; i <= n; i++) {
    if ((b[i] + c[i] - sum) % n != 0) {
      puts("-1");
      return 0;
    }
    a[i] = (b[i] + c[i] - sum) / n;
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; (1 << j) <= a[i]; j++) {
      if (a[i] & (1 << j)) cnt[j]++;
    }
  }
  for (i = 1; i <= n; i++) {
    long long b1 = 0, c1 = sum;
    for (j = 0; (1 << j) <= a[i]; j++) {
      if (a[i] & (1 << j))
        b1 += (1 << j) * cnt[j], c1 += (1 << j) * (n - cnt[j]);
    }
    if (b1 != b[i] || c1 != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (i = 1; i <= n; i++) printf("%lld ", a[i]);
  puts("");
  return 0;
}

```