#include <bits/stdc++.h>
const int inf = (int)1e9;
using namespace std;
int n;
long long a[200011], b[200011], sum = 0, x;
int f[200011];
void get(int x) {
  int dva = 1073741824;
  int h = 30;
  while (dva) {
    if (x >= dva) {
      f[h]++;
      x -= dva;
    }
    dva /= 2;
    h--;
  }
}
void check(int x, int i) {
  int dva = 1073741824;
  int h = 30;
  long long an = 0, y = 0;
  while (dva) {
    if (x >= dva) {
      y += dva * n;
      an += dva * f[h];
      x -= dva;
    } else
      y += dva * f[h];
    dva /= 2;
    h--;
  }
  if (an != a[i] || y != b[i]) {
    cout << -1;
    exit(0);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    sum += b[i];
  }
  if (sum % (n * 2) != 0) {
    cout << -1;
    return 0;
  }
  x = sum / (n * 2);
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    a[i] += b[i];
    a[i] -= x;
    if (a[i] % n != 0 || a[i] < 0) {
      cout << -1;
      return 0;
    }
    ans.push_back(a[i] / n);
    a[i] += x;
    a[i] -= b[i];
    get(ans[i - 1]);
  }
  for (int i = 1; i <= n; i++) {
    check(ans[i - 1], i);
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i - 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>
const int inf = (int)1e9;
using namespace std;
int n;
long long a[200011], b[200011], sum = 0, x;
int f[200011];
void get(int x) {
  int dva = 1073741824;
  int h = 30;
  while (dva) {
    if (x >= dva) {
      f[h]++;
      x -= dva;
    }
    dva /= 2;
    h--;
  }
}
void check(int x, int i) {
  int dva = 1073741824;
  int h = 30;
  long long an = 0, y = 0;
  while (dva) {
    if (x >= dva) {
      y += dva * n;
      an += dva * f[h];
      x -= dva;
    } else
      y += dva * f[h];
    dva /= 2;
    h--;
  }
  if (an != a[i] || y != b[i]) {
    cout << -1;
    exit(0);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    sum += b[i];
  }
  if (sum % (n * 2) != 0) {
    cout << -1;
    return 0;
  }
  x = sum / (n * 2);
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    a[i] += b[i];
    a[i] -= x;
    if (a[i] % n != 0 || a[i] < 0) {
      cout << -1;
      return 0;
    }
    ans.push_back(a[i] / n);
    a[i] += x;
    a[i] -= b[i];
    get(ans[i - 1]);
  }
  for (int i = 1; i <= n; i++) {
    check(ans[i - 1], i);
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i - 1] << " ";
  }
  return 0;
}

```