#include <bits/stdc++.h>
using namespace std;
string itos(int x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
long long dx[200010], tt, ttd, res[200010];
int n, a[200010], b[200010];
int bit[32];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  tt = accumulate(b, b + n, accumulate(a, a + n, 0LL));
  if (tt % (2 * n) || (n == 1 && (a[0] != b[0]))) {
    cout << -1 << endl;
    return 0;
  }
  tt /= (2 * n);
  for (int i = 0; i < n; i++) {
    dx[i] = (a[i] + b[i]) - (a[0] + b[0]);
    if (abs(dx[i]) % n) {
      cout << -1 << endl;
      return 0;
    }
    dx[i] /= n;
  }
  ttd = accumulate(dx, dx + n, 0LL);
  res[0] = (tt - ttd) / n;
  for (int i = 0; i < n; i++) res[i] = res[0] + dx[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 32; j++)
      if (res[i] & (1 << j)) bit[j]++;
  for (int i = 0; i < n; i++) {
    long long r1, r2;
    r1 = r2 = 0;
    for (int j = 0; j < 32; j++) {
      if (res[i] & (1 << j)) {
        r1 += (1LL << j) * bit[j];
        r2 += (1LL << j) * n;
      } else
        r2 += (1LL << j) * bit[j];
    }
    if (r1 != a[i] || r2 != b[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) cout << res[i] << (i + 1 == n ? '\n' : ' ');
  return 0;
}

### Prompt

Please formulate 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;
string itos(int x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
long long dx[200010], tt, ttd, res[200010];
int n, a[200010], b[200010];
int bit[32];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  tt = accumulate(b, b + n, accumulate(a, a + n, 0LL));
  if (tt % (2 * n) || (n == 1 && (a[0] != b[0]))) {
    cout << -1 << endl;
    return 0;
  }
  tt /= (2 * n);
  for (int i = 0; i < n; i++) {
    dx[i] = (a[i] + b[i]) - (a[0] + b[0]);
    if (abs(dx[i]) % n) {
      cout << -1 << endl;
      return 0;
    }
    dx[i] /= n;
  }
  ttd = accumulate(dx, dx + n, 0LL);
  res[0] = (tt - ttd) / n;
  for (int i = 0; i < n; i++) res[i] = res[0] + dx[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 32; j++)
      if (res[i] & (1 << j)) bit[j]++;
  for (int i = 0; i < n; i++) {
    long long r1, r2;
    r1 = r2 = 0;
    for (int j = 0; j < 32; j++) {
      if (res[i] & (1 << j)) {
        r1 += (1LL << j) * bit[j];
        r2 += (1LL << j) * n;
      } else
        r2 += (1LL << j) * bit[j];
    }
    if (r1 != a[i] || r2 != b[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) cout << res[i] << (i + 1 == n ? '\n' : ' ');
  return 0;
}

```