#include <bits/stdc++.h>
using namespace std;
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
  for (Tp current = _begin; current != _end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<const int>(1e9 + 7);
constexpr int MAXN = 200000;
int arr[MAXN];
int brr[MAXN];
int ans[MAXN];
ll cnt[30];
bool HasBit(int x, int b) { return (x >> b) & 1; }
bool Check(int n) {
  for (int i = 0; i < (n); ++i) {
    for (int b = 0; b < 30; ++b) {
      if (HasBit(ans[i], b)) {
        ++cnt[b];
      }
    }
  }
  for (int i = 0; i < (n); ++i) {
    ll res = 0;
    for (int b = 0; b < 30; ++b) {
      if (HasBit(ans[i], b)) {
        res += cnt[b] << b;
      }
    }
    if (arr[i] != res) {
      return false;
    }
  }
  return true;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    cin >> arr[i];
  }
  for (int i = 0; i < (n); ++i) {
    cin >> brr[i];
  }
  for (int i = 0; i < (n); ++i) {
    if (arr[i] > brr[i]) {
      return cout << (-1) << '\n', 0;
      ;
    }
  }
  ll sum = accumulate(arr, arr + n, 0LL) + accumulate(brr, brr + n, 0LL);
  if (sum % (n + n) != 0) {
    return cout << (-1) << '\n', 0;
    ;
  }
  sum /= n + n;
  for (int i = 0; i < (n); ++i) {
    ans[i] = int(arr[i] + brr[i] - sum);
    if (ans[i] % n != 0) {
      return cout << (-1) << '\n', 0;
      ;
    }
    ans[i] /= n;
  }
  if (!Check(n)) {
    return cout << (-1) << '\n', 0;
    ;
  }
  outarr(ans, ans + n);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
  for (Tp current = _begin; current != _end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<const int>(1e9 + 7);
constexpr int MAXN = 200000;
int arr[MAXN];
int brr[MAXN];
int ans[MAXN];
ll cnt[30];
bool HasBit(int x, int b) { return (x >> b) & 1; }
bool Check(int n) {
  for (int i = 0; i < (n); ++i) {
    for (int b = 0; b < 30; ++b) {
      if (HasBit(ans[i], b)) {
        ++cnt[b];
      }
    }
  }
  for (int i = 0; i < (n); ++i) {
    ll res = 0;
    for (int b = 0; b < 30; ++b) {
      if (HasBit(ans[i], b)) {
        res += cnt[b] << b;
      }
    }
    if (arr[i] != res) {
      return false;
    }
  }
  return true;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    cin >> arr[i];
  }
  for (int i = 0; i < (n); ++i) {
    cin >> brr[i];
  }
  for (int i = 0; i < (n); ++i) {
    if (arr[i] > brr[i]) {
      return cout << (-1) << '\n', 0;
      ;
    }
  }
  ll sum = accumulate(arr, arr + n, 0LL) + accumulate(brr, brr + n, 0LL);
  if (sum % (n + n) != 0) {
    return cout << (-1) << '\n', 0;
    ;
  }
  sum /= n + n;
  for (int i = 0; i < (n); ++i) {
    ans[i] = int(arr[i] + brr[i] - sum);
    if (ans[i] % n != 0) {
      return cout << (-1) << '\n', 0;
      ;
    }
    ans[i] /= n;
  }
  if (!Check(n)) {
    return cout << (-1) << '\n', 0;
    ;
  }
  outarr(ans, ans + n);
  return 0;
}

```