#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
const int M = 63;
template <class T>
inline T rd() {
  T x = 0;
  bool f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = 0;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  return f ? x : -x;
}
inline void NO() {
  puts("-1");
  exit(0);
}
int n;
long long a[N], b[N], c[N], s;
long long bit[M], u[N], v[N];
int main() {
  n = rd<int>();
  for (int i = (1), _ = (n); i <= _; ++i) b[i] = rd<long long>();
  for (int i = (1), _ = (n); i <= _; ++i) c[i] = rd<long long>();
  for (int i = (1), _ = (n); i <= _; ++i) s += b[i] + c[i];
  if (s % (2 * n)) NO();
  s /= 2 * n;
  for (int i = (1), _ = (n); i <= _; ++i) {
    long long d = b[i] + c[i] - s;
    if (d < 0 || d % n) NO();
    a[i] = d / n;
  }
  memset(bit, 0, sizeof bit);
  for (int i = (1), _ = (n); i <= _; ++i) {
    for (int j = (0), _ = (60); j <= _; ++j)
      if (a[i] >> j & 1) bit[j]++;
    for (int j = (0), _ = (60); j <= _; ++j) {
      if (a[i] >> j & 1) {
        u[i] += bit[j] * (1LL << j);
        v[i] += i * (1LL << j);
      } else {
        u[i] += 0;
        v[i] += bit[j] * (1LL << j);
      }
    }
  }
  memset(bit, 0, sizeof bit);
  for (int i = (n), _ = (1); i >= _; --i) {
    for (int j = (0), _ = (60); j <= _; ++j) {
      if (a[i] >> j & 1) {
        u[i] += bit[j] * (1LL << j);
        v[i] += (n - i) * (1LL << j);
      } else {
        u[i] += 0;
        v[i] += bit[j] * (1LL << j);
      }
    }
    for (int j = (0), _ = (60); j <= _; ++j)
      if (a[i] >> j & 1) bit[j]++;
  }
  for (int i = (1), _ = (n); i <= _; ++i)
    if (u[i] != b[i] || v[i] != c[i]) NO();
  for (int i = (1), _ = (n); i <= _; ++i)
    printf("%I64d%c", a[i], " \n"[i == n]);
  return 0;
}

### Prompt

Please formulate 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 int N = 2e5 + 7;
const int M = 63;
template <class T>
inline T rd() {
  T x = 0;
  bool f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = 0;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  return f ? x : -x;
}
inline void NO() {
  puts("-1");
  exit(0);
}
int n;
long long a[N], b[N], c[N], s;
long long bit[M], u[N], v[N];
int main() {
  n = rd<int>();
  for (int i = (1), _ = (n); i <= _; ++i) b[i] = rd<long long>();
  for (int i = (1), _ = (n); i <= _; ++i) c[i] = rd<long long>();
  for (int i = (1), _ = (n); i <= _; ++i) s += b[i] + c[i];
  if (s % (2 * n)) NO();
  s /= 2 * n;
  for (int i = (1), _ = (n); i <= _; ++i) {
    long long d = b[i] + c[i] - s;
    if (d < 0 || d % n) NO();
    a[i] = d / n;
  }
  memset(bit, 0, sizeof bit);
  for (int i = (1), _ = (n); i <= _; ++i) {
    for (int j = (0), _ = (60); j <= _; ++j)
      if (a[i] >> j & 1) bit[j]++;
    for (int j = (0), _ = (60); j <= _; ++j) {
      if (a[i] >> j & 1) {
        u[i] += bit[j] * (1LL << j);
        v[i] += i * (1LL << j);
      } else {
        u[i] += 0;
        v[i] += bit[j] * (1LL << j);
      }
    }
  }
  memset(bit, 0, sizeof bit);
  for (int i = (n), _ = (1); i >= _; --i) {
    for (int j = (0), _ = (60); j <= _; ++j) {
      if (a[i] >> j & 1) {
        u[i] += bit[j] * (1LL << j);
        v[i] += (n - i) * (1LL << j);
      } else {
        u[i] += 0;
        v[i] += bit[j] * (1LL << j);
      }
    }
    for (int j = (0), _ = (60); j <= _; ++j)
      if (a[i] >> j & 1) bit[j]++;
  }
  for (int i = (1), _ = (n); i <= _; ++i)
    if (u[i] != b[i] || v[i] != c[i]) NO();
  for (int i = (1), _ = (n); i <= _; ++i)
    printf("%I64d%c", a[i], " \n"[i == n]);
  return 0;
}

```