#include <bits/stdc++.h>
using namespace std;
long long N;
long long A[200005];
long long B[200005];
long long C[200005];
long long D[200005];
long long E[200005];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N;
  for (long long i = 0; i < N; ++i) {
    cin >> B[i];
  }
  for (long long i = 0; i < N; ++i) {
    cin >> C[i];
  }
  long long s = 0;
  for (long long i = 0; i < N; ++i) {
    s += B[i];
    s += C[i];
  }
  s = s / (2 * N);
  for (long long i = 0; i < N; ++i) {
    A[i] = (B[i] + C[i] - s) / N;
  }
  vector<long long> v;
  v.resize(64);
  for (long long i = 0; i < N; ++i) {
    for (long long j = 0; j < 64; ++j) {
      if (A[i] >> j & 1) {
        v[j]++;
      }
    }
  }
  for (long long i = 0; i < N; ++i) {
    long long b = 0;
    long long c = 0;
    for (long long j = 0; j < 64; ++j) {
      if ((A[i] >> j) & 1) {
        b += (1LL << j) * v[j];
        c += (1LL << j) * N;
      } else {
        c += (1LL << j) * v[j];
      }
    }
    if (b != B[i] || c != C[i]) {
      printf("-1\n");
      return 0;
    }
  }
  for (long long i = 0; i < N; ++i) {
    printf("%lld%c", A[i], i == N - 1 ? '\n' : ' ');
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long N;
long long A[200005];
long long B[200005];
long long C[200005];
long long D[200005];
long long E[200005];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N;
  for (long long i = 0; i < N; ++i) {
    cin >> B[i];
  }
  for (long long i = 0; i < N; ++i) {
    cin >> C[i];
  }
  long long s = 0;
  for (long long i = 0; i < N; ++i) {
    s += B[i];
    s += C[i];
  }
  s = s / (2 * N);
  for (long long i = 0; i < N; ++i) {
    A[i] = (B[i] + C[i] - s) / N;
  }
  vector<long long> v;
  v.resize(64);
  for (long long i = 0; i < N; ++i) {
    for (long long j = 0; j < 64; ++j) {
      if (A[i] >> j & 1) {
        v[j]++;
      }
    }
  }
  for (long long i = 0; i < N; ++i) {
    long long b = 0;
    long long c = 0;
    for (long long j = 0; j < 64; ++j) {
      if ((A[i] >> j) & 1) {
        b += (1LL << j) * v[j];
        c += (1LL << j) * N;
      } else {
        c += (1LL << j) * v[j];
      }
    }
    if (b != B[i] || c != C[i]) {
      printf("-1\n");
      return 0;
    }
  }
  for (long long i = 0; i < N; ++i) {
    printf("%lld%c", A[i], i == N - 1 ? '\n' : ' ');
  }
  return 0;
}

```