#include <bits/stdc++.h>
static constexpr int MAXN = 200010;
long long sums[MAXN];
long long B[MAXN];
long long C[MAXN];
long long ans[MAXN];
long long B_check[MAXN];
long long C_check[MAXN];
const int LOG = 63;
int main() {
  int N;
  std::cin >> N;
  for (int i = 0; i < N; i++) std::cin >> B[i];
  for (int i = 0; i < N; i++) {
    std::cin >> C[i];
    sums[i] = B[i] + C[i];
  }
  long long total = 0;
  for (int i = 0; i < N; i++) total += sums[i];
  total /= (2LL * N);
  for (int i = 0; i < N; i++) {
    ans[i] = (sums[i] - total) / N;
    if (ans[i] < 0) {
      std::cout << "-1\n";
      return 0;
    }
  }
  int cnt[LOG];
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < LOG; j++) {
      if (ans[i] & (1LL << j)) {
        cnt[j]++;
      }
    }
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < LOG; j++) {
      if (ans[i] & (1LL << j)) {
        B_check[i] += (1LL << j) * cnt[j];
        C_check[i] += (1LL << j) * N;
      } else {
        C_check[i] += (1LL << j) * cnt[j];
      }
    }
  }
  for (int i = 0; i < N; i++) {
    if (B_check[i] != B[i] || C_check[i] != C[i]) {
      std::cout << "-1\n";
      return 0;
    }
  }
  for (int i = 0; i < N; i++) std::cout << ans[i] << ' ';
  std::cout << '\n';
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
static constexpr int MAXN = 200010;
long long sums[MAXN];
long long B[MAXN];
long long C[MAXN];
long long ans[MAXN];
long long B_check[MAXN];
long long C_check[MAXN];
const int LOG = 63;
int main() {
  int N;
  std::cin >> N;
  for (int i = 0; i < N; i++) std::cin >> B[i];
  for (int i = 0; i < N; i++) {
    std::cin >> C[i];
    sums[i] = B[i] + C[i];
  }
  long long total = 0;
  for (int i = 0; i < N; i++) total += sums[i];
  total /= (2LL * N);
  for (int i = 0; i < N; i++) {
    ans[i] = (sums[i] - total) / N;
    if (ans[i] < 0) {
      std::cout << "-1\n";
      return 0;
    }
  }
  int cnt[LOG];
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < LOG; j++) {
      if (ans[i] & (1LL << j)) {
        cnt[j]++;
      }
    }
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < LOG; j++) {
      if (ans[i] & (1LL << j)) {
        B_check[i] += (1LL << j) * cnt[j];
        C_check[i] += (1LL << j) * N;
      } else {
        C_check[i] += (1LL << j) * cnt[j];
      }
    }
  }
  for (int i = 0; i < N; i++) {
    if (B_check[i] != B[i] || C_check[i] != C[i]) {
      std::cout << "-1\n";
      return 0;
    }
  }
  for (int i = 0; i < N; i++) std::cout << ans[i] << ' ';
  std::cout << '\n';
  return 0;
}

```