#include <bits/stdc++.h>
const int N = 200005;
int n;
long long sum, A[N], B[N], C[N], bin[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &A[i]), B[i] = A[i], sum += A[i];
  for (int i = 1; i <= n; i++) scanf("%I64d", &C[i]), A[i] += C[i], sum += C[i];
  if (sum % (2 * n)) return puts("-1"), 0;
  sum /= (2 * n);
  for (int i = 1; i <= n; i++)
    if (A[i] < sum || (A[i] - sum) % n) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) A[i] = (A[i] - sum) / n;
  for (int i = 30; i >= 0; i--)
    for (int j = 1; j <= n; j++)
      if (A[j] >> i & 1) bin[i]++;
  for (int j = 1; j <= n; j++) {
    long long Res = 0;
    for (int i = 0; i <= 30; i++)
      if (A[j] >> i & 1) Res += 1ll * bin[i] * (1 << i);
    if (B[j] != Res || 1ll * sum + A[j] * n - B[j] != C[j])
      return puts("-1"), 0;
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", A[i]);
  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>
const int N = 200005;
int n;
long long sum, A[N], B[N], C[N], bin[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &A[i]), B[i] = A[i], sum += A[i];
  for (int i = 1; i <= n; i++) scanf("%I64d", &C[i]), A[i] += C[i], sum += C[i];
  if (sum % (2 * n)) return puts("-1"), 0;
  sum /= (2 * n);
  for (int i = 1; i <= n; i++)
    if (A[i] < sum || (A[i] - sum) % n) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) A[i] = (A[i] - sum) / n;
  for (int i = 30; i >= 0; i--)
    for (int j = 1; j <= n; j++)
      if (A[j] >> i & 1) bin[i]++;
  for (int j = 1; j <= n; j++) {
    long long Res = 0;
    for (int i = 0; i <= 30; i++)
      if (A[j] >> i & 1) Res += 1ll * bin[i] * (1 << i);
    if (B[j] != Res || 1ll * sum + A[j] * n - B[j] != C[j])
      return puts("-1"), 0;
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", A[i]);
  return 0;
}

```