#include <bits/stdc++.h>
using namespace std;
long long a[200001], b[200001], c[200001], d[200001], s, n, cnt[200001];
int bit(int a, int j) { return (a & (1 << j)) >> j; }
int checkb() {
  int mb = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 29; ++j) {
      int x = bit(a[i], j), y = bit(b[i], j);
      if (x != 0) mb += cnt[j];
      if (y != mb % 2) return 0;
      mb /= 2;
    }
  }
  return 1;
}
int checkc() {
  int mc = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 29; ++j) {
      int x = bit(a[i], j), y = bit(c[i], j);
      if (x != 1)
        mc += cnt[j];
      else
        mc += n;
      if (y != mc % 2) return 0;
      mc /= 2;
    }
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  scanf("%I64d", &n);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &b[i]);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &c[i]);
  for (int i = 1; i <= n; ++i) d[i] = b[i] + c[i], s += d[i];
  s = s / (2 * n);
  for (int i = 1; i <= n; ++i) a[i] = (d[i] - s) / n;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 29; ++j) {
      int x = bit(a[i], j);
      if (x) cnt[j]++;
    }
  }
  if (checkb() && checkc())
    for (int i = 1; i <= n; ++i) printf("%I64d ", a[i]);
  else
    printf("-1");
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
long long a[200001], b[200001], c[200001], d[200001], s, n, cnt[200001];
int bit(int a, int j) { return (a & (1 << j)) >> j; }
int checkb() {
  int mb = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 29; ++j) {
      int x = bit(a[i], j), y = bit(b[i], j);
      if (x != 0) mb += cnt[j];
      if (y != mb % 2) return 0;
      mb /= 2;
    }
  }
  return 1;
}
int checkc() {
  int mc = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 29; ++j) {
      int x = bit(a[i], j), y = bit(c[i], j);
      if (x != 1)
        mc += cnt[j];
      else
        mc += n;
      if (y != mc % 2) return 0;
      mc /= 2;
    }
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  scanf("%I64d", &n);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &b[i]);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &c[i]);
  for (int i = 1; i <= n; ++i) d[i] = b[i] + c[i], s += d[i];
  s = s / (2 * n);
  for (int i = 1; i <= n; ++i) a[i] = (d[i] - s) / n;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 29; ++j) {
      int x = bit(a[i], j);
      if (x) cnt[j]++;
    }
  }
  if (checkb() && checkc())
    for (int i = 1; i <= n; ++i) printf("%I64d ", a[i]);
  else
    printf("-1");
  return 0;
}

```