#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 1, maxb = 32;
const long long inf = 2e18, mod = 1e9 + 7;
long long n, a[maxn], b[maxn], c[maxn], cnt[maxb], cntb[maxn][maxb],
    cntc[maxn][maxb], sum;
long long calc(long long ind, int type) {
  long long ans = 0;
  for (int i = (0); i < (maxb); i++)
    ans += (1 << i) * (type == 1 ? cntb[ind][i] : cntc[ind][i]);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = (0); i < (n); i++) cin >> b[i];
  for (int i = (0); i < (n); i++) cin >> c[i];
  for (int i = (0); i < (n); i++) {
    if (b[i] > c[i]) {
      cout << -1 << endl;
      return 0;
    }
    sum += b[i] + c[i];
  }
  if (sum % (2 * n)) {
    cout << -1 << endl;
    return 0;
  }
  sum /= (2 * n);
  for (int i = (0); i < (n); i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] % n) {
      cout << -1 << endl;
      return 0;
    }
    a[i] /= n;
    for (int j = (0); j < (maxb); j++) cnt[j] += ((a[i] >> j) & 1);
  }
  for (int i = (0); i < (n); i++)
    for (int j = (0); j < (maxb); j++) {
      if ((a[i] >> j) & 1) {
        cntc[i][j] += n;
        cntb[i][j] += cnt[j];
      } else
        cntc[i][j] += cnt[j];
    }
  bool check = true;
  for (int i = (0); i < (n); i++) {
    check &= (b[i] == calc(i, 1));
    check &= (c[i] == calc(i, 2));
  }
  if (!check)
    cout << -1;
  else
    for (int i = (0); i < (n); i++) cout << a[i] << ' ';
  cout << endl;
  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>
using namespace std;
const int maxn = 2e5 + 1, maxb = 32;
const long long inf = 2e18, mod = 1e9 + 7;
long long n, a[maxn], b[maxn], c[maxn], cnt[maxb], cntb[maxn][maxb],
    cntc[maxn][maxb], sum;
long long calc(long long ind, int type) {
  long long ans = 0;
  for (int i = (0); i < (maxb); i++)
    ans += (1 << i) * (type == 1 ? cntb[ind][i] : cntc[ind][i]);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = (0); i < (n); i++) cin >> b[i];
  for (int i = (0); i < (n); i++) cin >> c[i];
  for (int i = (0); i < (n); i++) {
    if (b[i] > c[i]) {
      cout << -1 << endl;
      return 0;
    }
    sum += b[i] + c[i];
  }
  if (sum % (2 * n)) {
    cout << -1 << endl;
    return 0;
  }
  sum /= (2 * n);
  for (int i = (0); i < (n); i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] % n) {
      cout << -1 << endl;
      return 0;
    }
    a[i] /= n;
    for (int j = (0); j < (maxb); j++) cnt[j] += ((a[i] >> j) & 1);
  }
  for (int i = (0); i < (n); i++)
    for (int j = (0); j < (maxb); j++) {
      if ((a[i] >> j) & 1) {
        cntc[i][j] += n;
        cntb[i][j] += cnt[j];
      } else
        cntc[i][j] += cnt[j];
    }
  bool check = true;
  for (int i = (0); i < (n); i++) {
    check &= (b[i] == calc(i, 1));
    check &= (c[i] == calc(i, 2));
  }
  if (!check)
    cout << -1;
  else
    for (int i = (0); i < (n); i++) cout << a[i] << ' ';
  cout << endl;
  return 0;
}

```