#include <bits/stdc++.h>
using namespace std;
bool chkmin(int &a, int b) { return b < a ? a = b, 1 : 0; }
bool chkmax(int &a, int b) { return b > a ? a = b, 1 : 0; }
inline long long read() {
  long long x = 0, fh = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fh = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ '0');
  return x * fh;
}
int a[200005], b[200005];
int c[200005], n;
int num[40];
int main() {
  n = read();
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) b[i] = read();
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) c[i] = read();
  long long sum = 0;
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) sum += b[i] + c[i];
  if (sum % n || sum % 2) {
    puts("-1");
    return 0;
  }
  int flag = 1;
  sum /= 2 * n;
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) {
    int tmp = b[i] + c[i] - sum;
    if (tmp % n) {
      flag = 0;
      break;
    }
    a[i] = tmp / n;
  }
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) {
    for (int j = (1), _end_ = (int)(32); j <= _end_; ++j) {
      if ((a[i] >> (j - 1)) & 1) num[j]++;
    }
  }
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) {
    int sb = 0, sc = 0;
    for (int j = (1), _end_ = (int)(32); j <= _end_; ++j) {
      if ((a[i] >> (j - 1)) & 1) {
        sb += num[j] * (1 << (j - 1));
        sc += n * (1 << (j - 1));
      } else {
        sc += num[j] * (1 << (j - 1));
      }
    }
    if (sb != b[i] || sc != c[i]) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    for (int i = (1), _end_ = (int)(n); i <= _end_; ++i)
      printf("%d%c", a[i], i == n ? '\n' : ' ');
    return 0;
  }
  puts("-1");
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
bool chkmin(int &a, int b) { return b < a ? a = b, 1 : 0; }
bool chkmax(int &a, int b) { return b > a ? a = b, 1 : 0; }
inline long long read() {
  long long x = 0, fh = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fh = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ '0');
  return x * fh;
}
int a[200005], b[200005];
int c[200005], n;
int num[40];
int main() {
  n = read();
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) b[i] = read();
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) c[i] = read();
  long long sum = 0;
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) sum += b[i] + c[i];
  if (sum % n || sum % 2) {
    puts("-1");
    return 0;
  }
  int flag = 1;
  sum /= 2 * n;
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) {
    int tmp = b[i] + c[i] - sum;
    if (tmp % n) {
      flag = 0;
      break;
    }
    a[i] = tmp / n;
  }
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) {
    for (int j = (1), _end_ = (int)(32); j <= _end_; ++j) {
      if ((a[i] >> (j - 1)) & 1) num[j]++;
    }
  }
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) {
    int sb = 0, sc = 0;
    for (int j = (1), _end_ = (int)(32); j <= _end_; ++j) {
      if ((a[i] >> (j - 1)) & 1) {
        sb += num[j] * (1 << (j - 1));
        sc += n * (1 << (j - 1));
      } else {
        sc += num[j] * (1 << (j - 1));
      }
    }
    if (sb != b[i] || sc != c[i]) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    for (int i = (1), _end_ = (int)(n); i <= _end_; ++i)
      printf("%d%c", a[i], i == n ? '\n' : ' ');
    return 0;
  }
  puts("-1");
  return 0;
}

```