#include <bits/stdc++.h>
using namespace std;
const long long maxn = 3e5, N = 2e6 + 90, SQ = 300, base = 1999, mod = 1e9 + 7,
                INF = 1e9 + 1, lg = 30;
const long double eps = 1e-4;
long long n, a[maxn], b[maxn], c[maxn], d[maxn], cnt[maxn], s1, s2, tot;
inline bool check() {
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < lg; ++j) {
      if (a[i] & (1ll << j)) {
        cnt[j]++;
      }
    }
  }
  for (long long i = 0; i < n; ++i) {
    long long res1 = 0, res2 = 0;
    for (long long j = 0; j < lg; ++j) {
      if (a[i] & (1ll << j)) {
        res1 += (cnt[j] * (1ll << j));
        res2 += (n - cnt[j]) * (1ll << j);
      }
      res2 += (cnt[j] * (1ll << j));
    }
    if (res1 != b[i] || res2 != c[i]) {
      return 0;
    }
  }
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> b[i];
    s1 += b[i];
  }
  for (long long i = 0; i < n; ++i) {
    cin >> c[i];
    s2 += c[i];
  }
  if ((s1 + s2) % (2 * n) != 0) {
    cout << -1 << '\n';
    return 0;
  }
  tot = (s1 + s2) / (2 * n);
  for (long long i = 0; i < n; ++i) {
    d[i] = b[i] + c[i];
  }
  for (long long i = 0; i < n; ++i) {
    a[i] = (d[i] - tot) / n;
  }
  if (check()) {
    for (long long i = 0; i < n; ++i) {
      cout << a[i] << ' ';
    }
    cout << '\n';
  } else {
    cout << -1 << '\n';
  }
}

### 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;
const long long maxn = 3e5, N = 2e6 + 90, SQ = 300, base = 1999, mod = 1e9 + 7,
                INF = 1e9 + 1, lg = 30;
const long double eps = 1e-4;
long long n, a[maxn], b[maxn], c[maxn], d[maxn], cnt[maxn], s1, s2, tot;
inline bool check() {
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < lg; ++j) {
      if (a[i] & (1ll << j)) {
        cnt[j]++;
      }
    }
  }
  for (long long i = 0; i < n; ++i) {
    long long res1 = 0, res2 = 0;
    for (long long j = 0; j < lg; ++j) {
      if (a[i] & (1ll << j)) {
        res1 += (cnt[j] * (1ll << j));
        res2 += (n - cnt[j]) * (1ll << j);
      }
      res2 += (cnt[j] * (1ll << j));
    }
    if (res1 != b[i] || res2 != c[i]) {
      return 0;
    }
  }
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> b[i];
    s1 += b[i];
  }
  for (long long i = 0; i < n; ++i) {
    cin >> c[i];
    s2 += c[i];
  }
  if ((s1 + s2) % (2 * n) != 0) {
    cout << -1 << '\n';
    return 0;
  }
  tot = (s1 + s2) / (2 * n);
  for (long long i = 0; i < n; ++i) {
    d[i] = b[i] + c[i];
  }
  for (long long i = 0; i < n; ++i) {
    a[i] = (d[i] - tot) / n;
  }
  if (check()) {
    for (long long i = 0; i < n; ++i) {
      cout << a[i] << ' ';
    }
    cout << '\n';
  } else {
    cout << -1 << '\n';
  }
}

```