#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 4;
long long n, b[N], c[N], a[N], sum[30], power[30], cnt, tong, sum1;
vector<long long> vec;
long long getbit(long long a, long long b) { return (a >> b) & 1; }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) cin >> c[i];
  if (n == 1) {
    if (b[1] == c[1])
      return cout << b[1], 0;
    else
      return cout << -1, 0;
  }
  for (long long i = 1; i <= n; i++) {
    sum1 = sum1 + b[i] + c[i];
  }
  if (sum1 % (2 * n) != 0) return cout << -1, 0;
  sum1 /= (2 * n);
  power[0] = 1;
  for (long long i = 1; i <= 29; i++) power[i] = power[i - 1] * 2;
  for (long long i = 1; i <= n; i++) {
    if ((b[i] + c[i] - sum1) % n != 0) return cout << -1, 0;
    a[i] = (b[i] + c[i] - sum1) / n;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j <= 29; j++) {
      if (getbit(a[i], j) == 1) sum[j]++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    tong = 0;
    for (long long j = 0; j <= 29; j++) {
      cnt = 0;
      if (getbit(a[i], j) == 0)
        cnt = 0;
      else
        cnt = sum[j];
      tong = tong + power[j] * cnt;
    }
    if (tong != b[i]) return cout << -1, 0;
    tong = 0;
    for (long long j = 0; j <= 29; j++) {
      cnt = 0;
      if (getbit(a[i], j) == 0)
        cnt = sum[j];
      else
        cnt = n;
      tong = tong + power[j] * cnt;
    }
    if (tong != c[i]) return cout << -1, 0;
  }
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
}

### Prompt

Your task is to create 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;
const long long N = 2e5 + 4;
long long n, b[N], c[N], a[N], sum[30], power[30], cnt, tong, sum1;
vector<long long> vec;
long long getbit(long long a, long long b) { return (a >> b) & 1; }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) cin >> c[i];
  if (n == 1) {
    if (b[1] == c[1])
      return cout << b[1], 0;
    else
      return cout << -1, 0;
  }
  for (long long i = 1; i <= n; i++) {
    sum1 = sum1 + b[i] + c[i];
  }
  if (sum1 % (2 * n) != 0) return cout << -1, 0;
  sum1 /= (2 * n);
  power[0] = 1;
  for (long long i = 1; i <= 29; i++) power[i] = power[i - 1] * 2;
  for (long long i = 1; i <= n; i++) {
    if ((b[i] + c[i] - sum1) % n != 0) return cout << -1, 0;
    a[i] = (b[i] + c[i] - sum1) / n;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j <= 29; j++) {
      if (getbit(a[i], j) == 1) sum[j]++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    tong = 0;
    for (long long j = 0; j <= 29; j++) {
      cnt = 0;
      if (getbit(a[i], j) == 0)
        cnt = 0;
      else
        cnt = sum[j];
      tong = tong + power[j] * cnt;
    }
    if (tong != b[i]) return cout << -1, 0;
    tong = 0;
    for (long long j = 0; j <= 29; j++) {
      cnt = 0;
      if (getbit(a[i], j) == 0)
        cnt = sum[j];
      else
        cnt = n;
      tong = tong + power[j] * cnt;
    }
    if (tong != c[i]) return cout << -1, 0;
  }
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
}

```