#include <bits/stdc++.h>
using namespace std;
const long long maxn = 300909;
const long long maxN = 2e6 + 5;
const long long oo = 1e18;
const long long mod = 1e9 + 7;
long long n, a[maxn], b[maxn], c[maxn], d[maxn], b1[maxn], c1[maxn], s = 0;
long long e[40];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) {
    cin >> c[i];
    d[i] = c[i] + b[i];
    s += d[i];
  }
  s = s / (2 * n);
  for (long long i = 1; i <= n; i++) {
    a[i] = (d[i] - s) / n;
  }
  for (long long i = 0; i <= 30; i++) {
    for (long long j = 1; j <= n; j++) {
      e[i] += ((a[j] & (1 << i)) != 0);
    }
  }
  for (long long i = 0; i <= 30; i++) {
    for (long long j = 1; j <= n; j++) {
      if ((a[j] & (1 << i)) != 0) {
        b1[j] += ((1 << i) * e[i]);
        c1[j] += ((1 << i) * n);
      } else
        c1[j] += ((1 << i) * e[i]);
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (b[i] == b1[i] && c[i] == c1[i]) continue;
    cout << "-1\n";
    return 0;
  }
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
}

### 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;
const long long maxn = 300909;
const long long maxN = 2e6 + 5;
const long long oo = 1e18;
const long long mod = 1e9 + 7;
long long n, a[maxn], b[maxn], c[maxn], d[maxn], b1[maxn], c1[maxn], s = 0;
long long e[40];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) {
    cin >> c[i];
    d[i] = c[i] + b[i];
    s += d[i];
  }
  s = s / (2 * n);
  for (long long i = 1; i <= n; i++) {
    a[i] = (d[i] - s) / n;
  }
  for (long long i = 0; i <= 30; i++) {
    for (long long j = 1; j <= n; j++) {
      e[i] += ((a[j] & (1 << i)) != 0);
    }
  }
  for (long long i = 0; i <= 30; i++) {
    for (long long j = 1; j <= n; j++) {
      if ((a[j] & (1 << i)) != 0) {
        b1[j] += ((1 << i) * e[i]);
        c1[j] += ((1 << i) * n);
      } else
        c1[j] += ((1 << i) * e[i]);
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (b[i] == b1[i] && c[i] == c1[i]) continue;
    cout << "-1\n";
    return 0;
  }
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
}

```