#include <bits/stdc++.h>
using namespace std;
int n;
long long b[200009];
long long c[200009];
long long tmp[200009];
long long d[53];
long long pow(int x, int y) {
  if (y == 0) return 1;
  return pow(x, y - 1) * x;
}
int main() {
  cin >> n;
  long long x = 0;
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    x = x + b[i];
  }
  for (int j = 1; j <= n; j++) {
    cin >> c[j];
    x = x + c[j];
  }
  if (x % (2 * n) == 0) {
    x = x / (2 * n);
    for (int i = 1; i <= n; i++) {
      if ((b[i] + c[i] - x) % n == 0) {
        tmp[i] = (b[i] + c[i] - x) / n;
      } else {
        cout << "-1" << endl;
        return 0;
      }
    }
  } else {
    cout << "-1" << endl;
    return 0;
  }
  for (int i = 1; i <= min(10, n); i++) {
    long long tam = 0;
    long long dem = 0;
    for (int j = 1; j <= n; j++) {
      tam = tam + (tmp[i] & tmp[j]);
      dem = dem + (tmp[i] | tmp[j]);
    }
    if (tam != b[i] || dem != c[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= min(10, n); i++) {
    long long tam = 0;
    long long dem = 0;
    for (int j = 1; j <= n; j++) {
      tam = tam + (tmp[n - i + 1] & tmp[j]);
      dem = dem + (tmp[n - i + 1] | tmp[j]);
    }
    if (tam != b[n - i + 1] || dem != c[n - i + 1]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= 50; i++) {
    for (int j = 1; j <= n; j++) {
      long long o = tmp[j] & ((long long)1 << (i - 1));
      if (o != 0) {
        long long ak = ((long long)1 << (i - 1));
        d[i]++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    long long sum = 0;
    for (int j = 1; j <= 50; j++) {
      int k = ((long long)1 << (j - 1)) & tmp[i];
      if (k != 0) {
        sum = sum + d[j] * ((long long)1 << (j - 1));
      }
    }
    if (sum != b[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long sum = 0;
    for (int j = 1; j <= 50; j++) {
      int k = ((long long)1 << (j - 1)) & tmp[i];
      if (k != 0) {
        sum = sum + n * ((long long)1 << (j - 1));
      } else {
        sum = sum + d[j] * ((long long)1 << (j - 1));
      }
    }
    if (sum != c[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << tmp[i] << ' ';
  }
  return 0;
}

### Prompt

Please formulate 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 namespace std;
int n;
long long b[200009];
long long c[200009];
long long tmp[200009];
long long d[53];
long long pow(int x, int y) {
  if (y == 0) return 1;
  return pow(x, y - 1) * x;
}
int main() {
  cin >> n;
  long long x = 0;
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    x = x + b[i];
  }
  for (int j = 1; j <= n; j++) {
    cin >> c[j];
    x = x + c[j];
  }
  if (x % (2 * n) == 0) {
    x = x / (2 * n);
    for (int i = 1; i <= n; i++) {
      if ((b[i] + c[i] - x) % n == 0) {
        tmp[i] = (b[i] + c[i] - x) / n;
      } else {
        cout << "-1" << endl;
        return 0;
      }
    }
  } else {
    cout << "-1" << endl;
    return 0;
  }
  for (int i = 1; i <= min(10, n); i++) {
    long long tam = 0;
    long long dem = 0;
    for (int j = 1; j <= n; j++) {
      tam = tam + (tmp[i] & tmp[j]);
      dem = dem + (tmp[i] | tmp[j]);
    }
    if (tam != b[i] || dem != c[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= min(10, n); i++) {
    long long tam = 0;
    long long dem = 0;
    for (int j = 1; j <= n; j++) {
      tam = tam + (tmp[n - i + 1] & tmp[j]);
      dem = dem + (tmp[n - i + 1] | tmp[j]);
    }
    if (tam != b[n - i + 1] || dem != c[n - i + 1]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= 50; i++) {
    for (int j = 1; j <= n; j++) {
      long long o = tmp[j] & ((long long)1 << (i - 1));
      if (o != 0) {
        long long ak = ((long long)1 << (i - 1));
        d[i]++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    long long sum = 0;
    for (int j = 1; j <= 50; j++) {
      int k = ((long long)1 << (j - 1)) & tmp[i];
      if (k != 0) {
        sum = sum + d[j] * ((long long)1 << (j - 1));
      }
    }
    if (sum != b[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long sum = 0;
    for (int j = 1; j <= 50; j++) {
      int k = ((long long)1 << (j - 1)) & tmp[i];
      if (k != 0) {
        sum = sum + n * ((long long)1 << (j - 1));
      } else {
        sum = sum + d[j] * ((long long)1 << (j - 1));
      }
    }
    if (sum != c[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << tmp[i] << ' ';
  }
  return 0;
}

```