#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int maxn = 2e5 + 20;
int n;
long long a[maxn], b[maxn], c[maxn], d[maxn];
int cnt[40];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]);
  for (int i = 1; i <= n; i++) scanf("%I64d", &c[i]);
  long long sum = 0;
  for (int i = 1; i <= n; i++) d[i] = c[i] + b[i], sum += d[i];
  if (sum % (2 * n) != 0) {
    printf("-1\n");
    return 0;
  }
  sum /= (2 * n);
  for (int i = 1; i <= n; i++)
    if ((d[i] - sum) % n != 0 || d[i] < sum) {
      printf("-1\n");
      return 0;
    } else
      a[i] = (d[i] - sum) / n;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 40; j++) cnt[j] += ((a[i] >> j) & 1);
  for (int i = 1; i <= n; i++) {
    long long x = 0, y = 0;
    for (int j = 0; j < 40; j++)
      if ((a[i] >> j) & 1) {
        x += (1ll << j) * cnt[j];
        y += (1ll << j) * n;
      } else
        y += cnt[j] * (1ll << j);
    if (x == b[i] && y == c[i]) continue;
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}

### 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;
const int INF = 1e9;
const int maxn = 2e5 + 20;
int n;
long long a[maxn], b[maxn], c[maxn], d[maxn];
int cnt[40];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]);
  for (int i = 1; i <= n; i++) scanf("%I64d", &c[i]);
  long long sum = 0;
  for (int i = 1; i <= n; i++) d[i] = c[i] + b[i], sum += d[i];
  if (sum % (2 * n) != 0) {
    printf("-1\n");
    return 0;
  }
  sum /= (2 * n);
  for (int i = 1; i <= n; i++)
    if ((d[i] - sum) % n != 0 || d[i] < sum) {
      printf("-1\n");
      return 0;
    } else
      a[i] = (d[i] - sum) / n;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 40; j++) cnt[j] += ((a[i] >> j) & 1);
  for (int i = 1; i <= n; i++) {
    long long x = 0, y = 0;
    for (int j = 0; j < 40; j++)
      if ((a[i] >> j) & 1) {
        x += (1ll << j) * cnt[j];
        y += (1ll << j) * n;
      } else
        y += cnt[j] * (1ll << j);
    if (x == b[i] && y == c[i]) continue;
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}

```