#include <bits/stdc++.h>
using namespace std;
int n;
unsigned long long a[200200], b[200200], c[200200];
map<int, int> cnt;
int main(void) {
  unsigned long long tmp, sum = 0, tb, tc, p2[35];
  int tmax = -1;
  cin >> n;
  for (register int i = 1; i <= n; i++) {
    cin >> b[i];
    sum += b[i];
  }
  for (register int i = 1; i <= n; i++) {
    cin >> c[i];
    if (c[i] < b[i]) {
      cout << -1;
      return 0;
    }
    sum += c[i];
  }
  if (sum % (2 * n) == 0)
    sum /= 2 * n;
  else {
    cout << -1;
    return 0;
  }
  for (register int i = 1; i <= n; i++) {
    tmp = b[i] + c[i] - sum;
    if (tmp % n == 0)
      tmp = a[i] = tmp / n;
    else {
      cout << -1;
      return 0;
    }
    int tnt = 0;
    while (tmp) {
      if (tmp & 1) cnt[tnt]++;
      tnt++;
      tmp >>= 1;
    }
    if (tnt > tmax) tmax = tnt;
  }
  for (register int i = 0; i <= tmax; i++) p2[i] = pow(2, i);
  for (register int i = 1; i <= n; i++) {
    tmp = a[i];
    tb = tc = 0;
    int tnt = 0;
    while (tmp) {
      if (tmp & 1) {
        tb += cnt[tnt] * p2[tnt];
        tc += n * p2[tnt];
      } else
        tc += cnt[tnt] * p2[tnt];
      tnt++;
      tmp >>= 1;
    }
    for (; tnt <= tmax; tnt++) tc += cnt[tnt] * p2[tnt];
    if (tb != b[i] || tc != c[i]) {
      cout << -1;
      return 0;
    }
  }
  for (register int i = 1; i <= n; i++) cout << a[i] << ' ';
  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;
int n;
unsigned long long a[200200], b[200200], c[200200];
map<int, int> cnt;
int main(void) {
  unsigned long long tmp, sum = 0, tb, tc, p2[35];
  int tmax = -1;
  cin >> n;
  for (register int i = 1; i <= n; i++) {
    cin >> b[i];
    sum += b[i];
  }
  for (register int i = 1; i <= n; i++) {
    cin >> c[i];
    if (c[i] < b[i]) {
      cout << -1;
      return 0;
    }
    sum += c[i];
  }
  if (sum % (2 * n) == 0)
    sum /= 2 * n;
  else {
    cout << -1;
    return 0;
  }
  for (register int i = 1; i <= n; i++) {
    tmp = b[i] + c[i] - sum;
    if (tmp % n == 0)
      tmp = a[i] = tmp / n;
    else {
      cout << -1;
      return 0;
    }
    int tnt = 0;
    while (tmp) {
      if (tmp & 1) cnt[tnt]++;
      tnt++;
      tmp >>= 1;
    }
    if (tnt > tmax) tmax = tnt;
  }
  for (register int i = 0; i <= tmax; i++) p2[i] = pow(2, i);
  for (register int i = 1; i <= n; i++) {
    tmp = a[i];
    tb = tc = 0;
    int tnt = 0;
    while (tmp) {
      if (tmp & 1) {
        tb += cnt[tnt] * p2[tnt];
        tc += n * p2[tnt];
      } else
        tc += cnt[tnt] * p2[tnt];
      tnt++;
      tmp >>= 1;
    }
    for (; tnt <= tmax; tnt++) tc += cnt[tnt] * p2[tnt];
    if (tb != b[i] || tc != c[i]) {
      cout << -1;
      return 0;
    }
  }
  for (register int i = 1; i <= n; i++) cout << a[i] << ' ';
  return 0;
}

```