#include <bits/stdc++.h>
using UIntType = unsigned long long int;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::size_t n{};
  std::cin >> n;
  std::vector<UIntType> b(n);
  for (auto& bi : b) {
    std::cin >> bi;
  }
  std::vector<UIntType> c(n);
  for (auto& ci : c) {
    std::cin >> ci;
  }
  UIntType sum{std::accumulate(b.begin(), b.end(), UIntType{}) +
               std::accumulate(c.begin(), c.end(), UIntType{})};
  if (sum % (2 * n) != 0) {
    std::cout << -1 << '\n';
    return 0;
  }
  UIntType s{sum / (2 * n)};
  std::vector<UIntType> a;
  a.reserve(n);
  for (std::size_t i = 0; i < n; ++i) {
    a.emplace_back((b[i] + c[i] - s) / n);
  }
  std::vector<UIntType> bitCount(64);
  for (UIntType i = 0; i < 64; ++i) {
    for (const auto& ai : a) {
      if (((ai >> i) & 1) != 0) {
        ++bitCount[i];
      }
    }
  }
  for (std::size_t i = 0; i < n; ++i) {
    UIntType bi{};
    UIntType ci{};
    for (UIntType j = 0; j < 64; ++j) {
      if (((a[i] >> j) & 1) != 0) {
        bi += (UIntType{1} << j) * bitCount[j];
        ci += (UIntType{1} << j) * n;
      } else {
        ci += (UIntType{1} << j) * bitCount[j];
      }
    }
    if (bi != b[i] || ci != c[i]) {
      std::cout << -1 << '\n';
      return 0;
    }
  }
  for (const auto& ai : a) {
    std::cout << ai << ' ';
  }
  std::cout << '\n';
  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 UIntType = unsigned long long int;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::size_t n{};
  std::cin >> n;
  std::vector<UIntType> b(n);
  for (auto& bi : b) {
    std::cin >> bi;
  }
  std::vector<UIntType> c(n);
  for (auto& ci : c) {
    std::cin >> ci;
  }
  UIntType sum{std::accumulate(b.begin(), b.end(), UIntType{}) +
               std::accumulate(c.begin(), c.end(), UIntType{})};
  if (sum % (2 * n) != 0) {
    std::cout << -1 << '\n';
    return 0;
  }
  UIntType s{sum / (2 * n)};
  std::vector<UIntType> a;
  a.reserve(n);
  for (std::size_t i = 0; i < n; ++i) {
    a.emplace_back((b[i] + c[i] - s) / n);
  }
  std::vector<UIntType> bitCount(64);
  for (UIntType i = 0; i < 64; ++i) {
    for (const auto& ai : a) {
      if (((ai >> i) & 1) != 0) {
        ++bitCount[i];
      }
    }
  }
  for (std::size_t i = 0; i < n; ++i) {
    UIntType bi{};
    UIntType ci{};
    for (UIntType j = 0; j < 64; ++j) {
      if (((a[i] >> j) & 1) != 0) {
        bi += (UIntType{1} << j) * bitCount[j];
        ci += (UIntType{1} << j) * n;
      } else {
        ci += (UIntType{1} << j) * bitCount[j];
      }
    }
    if (bi != b[i] || ci != c[i]) {
      std::cout << -1 << '\n';
      return 0;
    }
  }
  for (const auto& ai : a) {
    std::cout << ai << ' ';
  }
  std::cout << '\n';
  return 0;
}

```