#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void init(T& x) {
  x = 0;
  char ch = getchar();
  bool t = 0;
  for (; ch > '9' || ch < '0'; ch = getchar())
    if (ch == '-') t = 1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + (ch - 48);
  if (t) x = -x;
  return;
}
const int N = 2e5 + 10;
long long S = 0;
int n;
long long a[N], b[N], c[N], d[N];
long long nb[N], nc[N];
int main() {
  init(n);
  for (int i = 1; i <= n; ++i) init(b[i]);
  for (int i = 1; i <= n; ++i) init(c[i]);
  for (int i = 1; i <= n; ++i) {
    d[i] = b[i] + c[i];
    S += d[i];
  }
  if (S % (n << 1)) return puts("-1"), 0;
  S /= n << 1;
  for (int i = 1; i <= n; ++i) {
    if ((d[i] - S) % n) return puts("-1"), 0;
    a[i] = (d[i] - S) / n;
    if (a[i] < 0) return puts("-1"), 0;
  }
  for (int k = 0; k <= 30; ++k) {
    int cnt = 0;
    for (int i = 1; i <= n; ++i)
      if ((a[i] >> k) & 1) ++cnt;
    for (int i = 1; i <= n; ++i) {
      if ((a[i] >> k) & 1) {
        nb[i] += (1ll << k) * cnt;
        nc[i] += (1ll << k) * n;
      } else {
        nc[i] += (1ll << k) * cnt;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if ((b[i] ^ nb[i]) || (c[i] ^ nc[i])) return puts("-1"), 0;
  }
  for (int i = 1; i <= n; ++i) printf("%lld ", a[i]);
  puts("");
  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;
template <class T>
inline void init(T& x) {
  x = 0;
  char ch = getchar();
  bool t = 0;
  for (; ch > '9' || ch < '0'; ch = getchar())
    if (ch == '-') t = 1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + (ch - 48);
  if (t) x = -x;
  return;
}
const int N = 2e5 + 10;
long long S = 0;
int n;
long long a[N], b[N], c[N], d[N];
long long nb[N], nc[N];
int main() {
  init(n);
  for (int i = 1; i <= n; ++i) init(b[i]);
  for (int i = 1; i <= n; ++i) init(c[i]);
  for (int i = 1; i <= n; ++i) {
    d[i] = b[i] + c[i];
    S += d[i];
  }
  if (S % (n << 1)) return puts("-1"), 0;
  S /= n << 1;
  for (int i = 1; i <= n; ++i) {
    if ((d[i] - S) % n) return puts("-1"), 0;
    a[i] = (d[i] - S) / n;
    if (a[i] < 0) return puts("-1"), 0;
  }
  for (int k = 0; k <= 30; ++k) {
    int cnt = 0;
    for (int i = 1; i <= n; ++i)
      if ((a[i] >> k) & 1) ++cnt;
    for (int i = 1; i <= n; ++i) {
      if ((a[i] >> k) & 1) {
        nb[i] += (1ll << k) * cnt;
        nc[i] += (1ll << k) * n;
      } else {
        nc[i] += (1ll << k) * cnt;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if ((b[i] ^ nb[i]) || (c[i] ^ nc[i])) return puts("-1"), 0;
  }
  for (int i = 1; i <= n; ++i) printf("%lld ", a[i]);
  puts("");
  return 0;
}

```