#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 7;
int b[MAXN], c[MAXN], a[MAXN], n, cnt[40], B[MAXN], C[MAXN];
int getBit(int v, int p) { return (v >> p) & 1; }
int main() {
  if (fopen("tst.inp", "r")) {
    freopen("tst.inp", "r", stdin);
    freopen("tst.out", "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  long long s = 0;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> b[i], s += b[i];
  for (int i = 1; i <= n; ++i) cin >> c[i], s += c[i];
  s /= 2LL * n;
  for (int i = 1; i <= n; ++i) {
    if ((b[i] + c[i] - s) % n != 0 || b[i] + c[i] < s) return cout << -1, 0;
    a[i] = (b[i] + c[i] - s) / n;
  }
  for (int j = 30; j >= 0; --j)
    for (int i = 1; i <= n; ++i)
      if (getBit(a[i], j)) ++cnt[j];
  for (int i = 1; i <= n; ++i)
    for (int j = 30; j >= 0; --j)
      if (getBit(a[i], j))
        B[i] += cnt[j] << j, C[i] += n << j;
      else
        C[i] += cnt[j] << j;
  for (int i = 1; i <= n; ++i)
    if (B[i] != b[i] || C[i] != c[i]) return cout << -1, 0;
  for (int i = 1; i <= n; ++i) cout << a[i] << ' ';
  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;
const int MAXN = 2e5 + 7;
int b[MAXN], c[MAXN], a[MAXN], n, cnt[40], B[MAXN], C[MAXN];
int getBit(int v, int p) { return (v >> p) & 1; }
int main() {
  if (fopen("tst.inp", "r")) {
    freopen("tst.inp", "r", stdin);
    freopen("tst.out", "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  long long s = 0;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> b[i], s += b[i];
  for (int i = 1; i <= n; ++i) cin >> c[i], s += c[i];
  s /= 2LL * n;
  for (int i = 1; i <= n; ++i) {
    if ((b[i] + c[i] - s) % n != 0 || b[i] + c[i] < s) return cout << -1, 0;
    a[i] = (b[i] + c[i] - s) / n;
  }
  for (int j = 30; j >= 0; --j)
    for (int i = 1; i <= n; ++i)
      if (getBit(a[i], j)) ++cnt[j];
  for (int i = 1; i <= n; ++i)
    for (int j = 30; j >= 0; --j)
      if (getBit(a[i], j))
        B[i] += cnt[j] << j, C[i] += n << j;
      else
        C[i] += cnt[j] << j;
  for (int i = 1; i <= n; ++i)
    if (B[i] != b[i] || C[i] != c[i]) return cout << -1, 0;
  for (int i = 1; i <= n; ++i) cout << a[i] << ' ';
  return 0;
}

```