#include <bits/stdc++.h>
using namespace std;
const int N = int(3e5), mod = int(1e9) + 7;
int n, b[N], c[N];
long long second, a[N];
int x[50];
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++) {
    second += b[i] + c[i];
  }
  if (second % (n + n) != 0) {
    cout << -1;
    return 0;
  }
  second /= (n + n);
  for (int i = 1; i <= n; i++) {
    if (second > b[i] + c[i] || ((b[i] + c[i] - second) % n)) {
      cout << -1;
      return 0;
    }
    a[i] = (b[i] + c[i] - second) / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 40; j++) {
      if (a[i] & (1ll << j)) x[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long cc = 0, bb = 0;
    for (int j = 0; j < 40; j++) {
      if (a[i] & (1ll << j)) {
        bb += 1ll * x[j] * (1 << j);
        cc += 1ll * n * (1 << j);
      } else {
        cc += 1ll * x[j] * (1 << j);
      }
    }
    if (cc != c[i] || bb != b[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%I64d ", 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 N = int(3e5), mod = int(1e9) + 7;
int n, b[N], c[N];
long long second, a[N];
int x[50];
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++) {
    second += b[i] + c[i];
  }
  if (second % (n + n) != 0) {
    cout << -1;
    return 0;
  }
  second /= (n + n);
  for (int i = 1; i <= n; i++) {
    if (second > b[i] + c[i] || ((b[i] + c[i] - second) % n)) {
      cout << -1;
      return 0;
    }
    a[i] = (b[i] + c[i] - second) / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 40; j++) {
      if (a[i] & (1ll << j)) x[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long cc = 0, bb = 0;
    for (int j = 0; j < 40; j++) {
      if (a[i] & (1ll << j)) {
        bb += 1ll * x[j] * (1 << j);
        cc += 1ll * n * (1 << j);
      } else {
        cc += 1ll * x[j] * (1 << j);
      }
    }
    if (cc != c[i] || bb != b[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%I64d ", a[i]);
  }
  return 0;
}

```