#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 20;
int b[maxn];
int c[maxn];
int a[maxn];
int t[maxn];
int n;
bool dp[maxn][32];
int cnt[32];
bool check() {
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 30; ++j) {
      dp[i][j] = a[i] & (1 << j);
      cnt[j] += dp[i][j];
    }
  }
  for (int i = 1; i <= n; ++i) {
    int tb = 0;
    int tc = 0;
    for (int j = 0; j <= 30; ++j) {
      if (dp[i][j]) {
        tc += (1 << j) * n;
      } else {
        tc += (1 << j) * cnt[j];
      }
      if (dp[i][j] == 0) continue;
      tb += (1 << j) * cnt[j];
    }
    if (tb != b[i] || tc != c[i]) {
      return false;
    }
  }
  return true;
}
void work() {
  cin >> n;
  long long int sum = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> b[i];
    sum += b[i];
  }
  for (int i = 1; i <= n; ++i) {
    cin >> c[i];
    t[i] = b[i] + c[i];
    sum += c[i];
  }
  if (sum % (2 * n) != 0) {
    printf("-1\n");
    return;
  }
  long long int tsum = sum / (2 * n);
  for (int i = 1; i <= n; ++i) {
    if (t[i] - tsum < 0 || (t[i] - tsum) % n != 0) {
      printf("-1\n");
      return;
    }
    a[i] = (t[i] - tsum) / n;
  }
  if (!check()) {
    cout << "-1" << endl;
    return;
  }
  for (int i = 1; i <= n; ++i) {
    cout << a[i] << " ";
  }
}
int main() {
  work();
  return 0;
}

### Prompt

In cpp, your task is to solve 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 int maxn = 200000 + 20;
int b[maxn];
int c[maxn];
int a[maxn];
int t[maxn];
int n;
bool dp[maxn][32];
int cnt[32];
bool check() {
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 30; ++j) {
      dp[i][j] = a[i] & (1 << j);
      cnt[j] += dp[i][j];
    }
  }
  for (int i = 1; i <= n; ++i) {
    int tb = 0;
    int tc = 0;
    for (int j = 0; j <= 30; ++j) {
      if (dp[i][j]) {
        tc += (1 << j) * n;
      } else {
        tc += (1 << j) * cnt[j];
      }
      if (dp[i][j] == 0) continue;
      tb += (1 << j) * cnt[j];
    }
    if (tb != b[i] || tc != c[i]) {
      return false;
    }
  }
  return true;
}
void work() {
  cin >> n;
  long long int sum = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> b[i];
    sum += b[i];
  }
  for (int i = 1; i <= n; ++i) {
    cin >> c[i];
    t[i] = b[i] + c[i];
    sum += c[i];
  }
  if (sum % (2 * n) != 0) {
    printf("-1\n");
    return;
  }
  long long int tsum = sum / (2 * n);
  for (int i = 1; i <= n; ++i) {
    if (t[i] - tsum < 0 || (t[i] - tsum) % n != 0) {
      printf("-1\n");
      return;
    }
    a[i] = (t[i] - tsum) / n;
  }
  if (!check()) {
    cout << "-1" << endl;
    return;
  }
  for (int i = 1; i <= n; ++i) {
    cout << a[i] << " ";
  }
}
int main() {
  work();
  return 0;
}

```