#include <bits/stdc++.h>
int n, B[300005], C[300005], tot[300005];
long long sum, A[300005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &B[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &C[i]);
  for (int i = 1; i <= n; i++) A[i] = B[i] + C[i];
  for (int i = n; i > 1; i--)
    if ((A[i] - A[i - 1]) % n == 0)
      A[i] = (A[i] - A[i - 1]) / n;
    else
      return puts("-1"), 0;
  for (int i = 2, s = 0; i <= n; i++) s += A[i], sum += s;
  A[1] = A[1] - sum;
  if (A[1] % (2 * n) != 0 || A[1] < 0) return puts("-1"), 0;
  A[1] /= n << 1;
  for (int i = 2; i <= n; i++) A[i] += A[i - 1];
  for (int i = 1; i <= n; i++) {
    if (A[i] < 0) return puts("-1"), 0;
    for (int j = 0; j < 30; j++) tot[j] += A[i] >> j & 1;
  }
  for (int i = 1; i <= n; i++) {
    long long Or = 0, And = 0;
    for (int j = 0; j <= 30; j++)
      if (A[i] >> j & 1)
        Or += 1ll * n * (1 << j), And += 1ll * tot[j] * (1 << j);
      else
        Or += 1ll * tot[j] * (1 << j);
    if (Or != C[i] || And != B[i]) return puts("-1");
  }
  for (int i = 1; i <= n; i++) printf("%d ", A[i]);
  return puts("");
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
int n, B[300005], C[300005], tot[300005];
long long sum, A[300005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &B[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &C[i]);
  for (int i = 1; i <= n; i++) A[i] = B[i] + C[i];
  for (int i = n; i > 1; i--)
    if ((A[i] - A[i - 1]) % n == 0)
      A[i] = (A[i] - A[i - 1]) / n;
    else
      return puts("-1"), 0;
  for (int i = 2, s = 0; i <= n; i++) s += A[i], sum += s;
  A[1] = A[1] - sum;
  if (A[1] % (2 * n) != 0 || A[1] < 0) return puts("-1"), 0;
  A[1] /= n << 1;
  for (int i = 2; i <= n; i++) A[i] += A[i - 1];
  for (int i = 1; i <= n; i++) {
    if (A[i] < 0) return puts("-1"), 0;
    for (int j = 0; j < 30; j++) tot[j] += A[i] >> j & 1;
  }
  for (int i = 1; i <= n; i++) {
    long long Or = 0, And = 0;
    for (int j = 0; j <= 30; j++)
      if (A[i] >> j & 1)
        Or += 1ll * n * (1 << j), And += 1ll * tot[j] * (1 << j);
      else
        Or += 1ll * tot[j] * (1 << j);
    if (Or != C[i] || And != B[i]) return puts("-1");
  }
  for (int i = 1; i <= n; i++) printf("%d ", A[i]);
  return puts("");
}

```