#include <bits/stdc++.h>
using namespace std;
int n;
long long a[200010] = {};
long long b[200010] = {};
long long c[200000] = {};
long long t[40] = {};
long long s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &b[i]);
    s += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
    s += c[i];
  }
  if (s % (2 * n) == 0) {
    s /= (2 * n);
  } else {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    long long p = 0;
    p = b[i] + c[i];
    p -= s;
    if (p < 0 || p % n) {
      puts("-1");
      return 0;
    } else {
      a[i] = p / n;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long q = a[i];
    int num = 1;
    long long o = 1;
    while (q) {
      if (q % 2) t[num] += o;
      q /= 2;
      num++;
      o <<= 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long bb = 0;
    long long cc = 0;
    long long q = a[i];
    int num = 1;
    long long o = 1;
    while (num != 32) {
      if (q % 2) {
        cc += n * o;
        bb += t[num];
      } else
        cc += t[num];
      q /= 2;
      num++;
      o <<= 1;
    }
    if (bb != b[i] || cc != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
long long a[200010] = {};
long long b[200010] = {};
long long c[200000] = {};
long long t[40] = {};
long long s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &b[i]);
    s += b[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
    s += c[i];
  }
  if (s % (2 * n) == 0) {
    s /= (2 * n);
  } else {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    long long p = 0;
    p = b[i] + c[i];
    p -= s;
    if (p < 0 || p % n) {
      puts("-1");
      return 0;
    } else {
      a[i] = p / n;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long q = a[i];
    int num = 1;
    long long o = 1;
    while (q) {
      if (q % 2) t[num] += o;
      q /= 2;
      num++;
      o <<= 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long bb = 0;
    long long cc = 0;
    long long q = a[i];
    int num = 1;
    long long o = 1;
    while (num != 32) {
      if (q % 2) {
        cc += n * o;
        bb += t[num];
      } else
        cc += t[num];
      q /= 2;
      num++;
      o <<= 1;
    }
    if (bb != b[i] || cc != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
  return 0;
}

```