#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <class T, class U>
inline void Max(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void Min(T &a, U b) {
  if (a > b) a = b;
}
inline void add(int &a, int b) {
  a += b;
  while (a >= 1000000007) a -= 1000000007;
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = ans * (long long)a % 1000000007;
    a = (long long)a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int b[200010], c[200010], a[200010], d[200010];
int main() {
  int T, i, j, k, ca = 0, m = 0, n, K;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%d", &c[i]), c[i] -= b[i];
  for (int i = 0; i < n; i++)
    if (c[i] < 0) {
      puts("-1");
      return 0;
    }
  long long s = 0;
  for (int i = 0; i < n; i++) c[i] += b[i], d[i] = c[i] + b[i], s += d[i];
  s /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = (d[i] - s) / n;
    if (a[i] < 0) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < 31; i++) d[i] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 31; j++) {
      d[j] += (a[i] >> j & 1);
    }
  }
  for (int i = 0; i < n; i++) {
    long long B = 0, C = 0;
    for (int j = 0; j < 31; j++) {
      long long v = 1 << j;
      if (a[i] >> j & 1)
        C += n * v, B += d[j] * v;
      else
        C += d[j] * v;
    }
    if (B != b[i] || C != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", a[i]);
  puts("");
  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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <class T, class U>
inline void Max(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void Min(T &a, U b) {
  if (a > b) a = b;
}
inline void add(int &a, int b) {
  a += b;
  while (a >= 1000000007) a -= 1000000007;
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = ans * (long long)a % 1000000007;
    a = (long long)a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int b[200010], c[200010], a[200010], d[200010];
int main() {
  int T, i, j, k, ca = 0, m = 0, n, K;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%d", &c[i]), c[i] -= b[i];
  for (int i = 0; i < n; i++)
    if (c[i] < 0) {
      puts("-1");
      return 0;
    }
  long long s = 0;
  for (int i = 0; i < n; i++) c[i] += b[i], d[i] = c[i] + b[i], s += d[i];
  s /= 2 * n;
  for (int i = 0; i < n; i++) {
    a[i] = (d[i] - s) / n;
    if (a[i] < 0) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < 31; i++) d[i] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 31; j++) {
      d[j] += (a[i] >> j & 1);
    }
  }
  for (int i = 0; i < n; i++) {
    long long B = 0, C = 0;
    for (int j = 0; j < 31; j++) {
      long long v = 1 << j;
      if (a[i] >> j & 1)
        C += n * v, B += d[j] * v;
      else
        C += d[j] * v;
    }
    if (B != b[i] || C != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", a[i]);
  puts("");
  return 0;
}

```