#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long a[N], b[N], c[N], bit[30];
int main() {
  int n, i, x, j;
  cin >> n;
  long long tot = 0;
  for (i = 1; i <= n; ++i) scanf("%d", &x), b[i] = x, tot += x;
  for (i = 1; i <= n; ++i) {
    scanf("%d", &x);
    if (b[i] > x) {
      printf("-1\n");
      return 0;
    }
    c[i] = x;
    tot += x;
  }
  if (tot % (n * 2)) {
    printf("-1");
    return 0;
  }
  tot /= (n * 2);
  for (i = 1; i <= n; ++i) {
    a[i] = c[i] + b[i] - tot;
    if (a[i] < 0 || (a[i] % n)) {
      printf("-1\n");
      return 0;
    }
    a[i] /= n;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 0; j < 30; ++j)
      if (a[i] & (1 << j)) bit[j] += 1 << j;
  }
  bool is = 1;
  for (i = 1; i <= n; ++i) {
    tot = 0;
    for (j = 0; j < 30; ++j)
      if (a[i] & (1 << j)) tot += bit[j];
    if (tot != b[i]) is = 0;
    tot = 0;
    for (j = 0; j < 30; ++j)
      if (!(a[i] & (1 << j))) tot += bit[j];
    tot += 1LL * n * a[i];
    if (tot != c[i]) is = 0;
  }
  if (!is)
    printf("-1\n");
  else
    for (i = 1; i <= n; ++i) cout << a[i] << " ";
  cout << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 2e5 + 5;
long long a[N], b[N], c[N], bit[30];
int main() {
  int n, i, x, j;
  cin >> n;
  long long tot = 0;
  for (i = 1; i <= n; ++i) scanf("%d", &x), b[i] = x, tot += x;
  for (i = 1; i <= n; ++i) {
    scanf("%d", &x);
    if (b[i] > x) {
      printf("-1\n");
      return 0;
    }
    c[i] = x;
    tot += x;
  }
  if (tot % (n * 2)) {
    printf("-1");
    return 0;
  }
  tot /= (n * 2);
  for (i = 1; i <= n; ++i) {
    a[i] = c[i] + b[i] - tot;
    if (a[i] < 0 || (a[i] % n)) {
      printf("-1\n");
      return 0;
    }
    a[i] /= n;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 0; j < 30; ++j)
      if (a[i] & (1 << j)) bit[j] += 1 << j;
  }
  bool is = 1;
  for (i = 1; i <= n; ++i) {
    tot = 0;
    for (j = 0; j < 30; ++j)
      if (a[i] & (1 << j)) tot += bit[j];
    if (tot != b[i]) is = 0;
    tot = 0;
    for (j = 0; j < 30; ++j)
      if (!(a[i] & (1 << j))) tot += bit[j];
    tot += 1LL * n * a[i];
    if (tot != c[i]) is = 0;
  }
  if (!is)
    printf("-1\n");
  else
    for (i = 1; i <= n; ++i) cout << a[i] << " ";
  cout << endl;
}

```