#include <bits/stdc++.h>
using namespace std;
void get(int &v) {
  char c;
  while ((c = getchar()) != EOF && isdigit(c) == 0)
    ;
  v = c - '0';
  while ((c = getchar()) != EOF && isdigit(c))
    v = (v << 3) + (v << 1) + c - '0';
  return;
}
int n, a[200000 + 70], b[200000 + 70], c[200000 + 70], d[200000 + 70], bit[32];
bool chk() {
  int i, j;
  long long S = 0, B, C, t;
  for (i = 1; i <= n; i++) get(b[i]);
  for (i = 1; i <= n; i++) {
    get(c[i]);
    S += b[i] + c[i];
  }
  S /= (n << 1);
  for (i = 1; i <= n; i++) {
    a[i] = (b[i] + c[i] - S) / n;
    for (j = 0; j < 31; j++) bit[j] += ((a[i] >> j) & 1);
  }
  for (i = 1; i <= n; i++) {
    B = C = 0;
    for (j = 0; j < 31; j++) {
      t = bit[j] << j;
      if ((a[i] >> j) & 1) {
        B += t;
        C += n * (1 << j);
      } else
        C += t;
    }
    if ((b[i] ^ B) || (c[i] ^ C)) return 0;
  }
  return 1;
}
int main() {
  int i;
  scanf("%d", &n);
  if (chk()) {
    for (i = 1; i < n; i++) printf("%d ", a[i]);
    printf("%d", a[i]);
  } else
    printf("-1");
  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;
void get(int &v) {
  char c;
  while ((c = getchar()) != EOF && isdigit(c) == 0)
    ;
  v = c - '0';
  while ((c = getchar()) != EOF && isdigit(c))
    v = (v << 3) + (v << 1) + c - '0';
  return;
}
int n, a[200000 + 70], b[200000 + 70], c[200000 + 70], d[200000 + 70], bit[32];
bool chk() {
  int i, j;
  long long S = 0, B, C, t;
  for (i = 1; i <= n; i++) get(b[i]);
  for (i = 1; i <= n; i++) {
    get(c[i]);
    S += b[i] + c[i];
  }
  S /= (n << 1);
  for (i = 1; i <= n; i++) {
    a[i] = (b[i] + c[i] - S) / n;
    for (j = 0; j < 31; j++) bit[j] += ((a[i] >> j) & 1);
  }
  for (i = 1; i <= n; i++) {
    B = C = 0;
    for (j = 0; j < 31; j++) {
      t = bit[j] << j;
      if ((a[i] >> j) & 1) {
        B += t;
        C += n * (1 << j);
      } else
        C += t;
    }
    if ((b[i] ^ B) || (c[i] ^ C)) return 0;
  }
  return 1;
}
int main() {
  int i;
  scanf("%d", &n);
  if (chk()) {
    for (i = 1; i < n; i++) printf("%d ", a[i]);
    printf("%d", a[i]);
  } else
    printf("-1");
  return 0;
}

```