#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
long long b[maxn], c[maxn];
long long a[maxn];
int w[60];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> b[i];
  for (int i = 0; i < n; ++i) cin >> c[i];
  long long sum = 0;
  for (int i = 0; i < n; ++i) sum += (b[i] + c[i]);
  sum /= 2 * n;
  long long S = 0;
  for (int i = 0; i < n; ++i) {
    a[i] = (b[i] + c[i] - sum) / n;
  }
  int flag = 1;
  memset(w, 0, sizeof w);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 60; ++j) {
      if (a[i] & ((1LL * 1) << j)) w[j]++;
    }
  }
  for (int i = 0; i < n; ++i) {
    long long B = 0, C = 0;
    for (int j = 0; j < 60; ++j) {
      if (a[i] & ((1LL * 1) << j)) {
        B += w[j] * (1 << j);
        C += n * (1 << j);
      } else
        C += w[j] * (1 << j);
    }
    if (B != b[i] || C != c[i]) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    for (int i = 0; i < n; ++i) cout << a[i] << ' ';
  } else
    cout << -1 << endl;
  return 0;
}

### Prompt

Generate 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 maxn = 200005;
long long b[maxn], c[maxn];
long long a[maxn];
int w[60];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> b[i];
  for (int i = 0; i < n; ++i) cin >> c[i];
  long long sum = 0;
  for (int i = 0; i < n; ++i) sum += (b[i] + c[i]);
  sum /= 2 * n;
  long long S = 0;
  for (int i = 0; i < n; ++i) {
    a[i] = (b[i] + c[i] - sum) / n;
  }
  int flag = 1;
  memset(w, 0, sizeof w);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 60; ++j) {
      if (a[i] & ((1LL * 1) << j)) w[j]++;
    }
  }
  for (int i = 0; i < n; ++i) {
    long long B = 0, C = 0;
    for (int j = 0; j < 60; ++j) {
      if (a[i] & ((1LL * 1) << j)) {
        B += w[j] * (1 << j);
        C += n * (1 << j);
      } else
        C += w[j] * (1 << j);
    }
    if (B != b[i] || C != c[i]) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    for (int i = 0; i < n; ++i) cout << a[i] << ' ';
  } else
    cout << -1 << endl;
  return 0;
}

```