#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int n;
long long b[maxn], c[maxn], s, a[maxn], num[63];
bool check() {
  memset(num, 0, sizeof num);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 63; ++j)
      if ((a[i] >> j) & 1) ++num[j];
  for (int i = 0; i < n; ++i) {
    long long B = 0;
    for (int j = 0; j < 63; ++j)
      if ((a[i] >> j) & 1) B += (1 << j) * num[j];
    if (B != b[i]) return false;
  }
  for (int i = 0; i < n; ++i) {
    long long C = 0;
    for (int j = 0; j < 63; ++j)
      if ((a[i] >> j) & 1)
        C += (1 << j) * n;
      else
        C += (1 << j) * num[j];
    if (C != c[i]) return false;
  }
  return true;
}
void solve() {
  if (s % (n << 1)) {
    puts("-1");
    return;
  }
  s /= (n << 1);
  for (int i = 0; i < n; ++i) {
    if (b[i] + c[i] - s < 0 || (b[i] + c[i] - s) % n) {
      puts("-1");
      return;
    }
    a[i] = (b[i] + c[i] - s) / n;
  }
  if (!check()) {
    puts("-1");
    return;
  }
  for (int i = 0; i < n; ++i) printf("%lld%c", a[i], i + 1 == n ? '\n' : ' ');
}
int main() {
  while (scanf("%d", &n) == 1) {
    s = 0;
    for (int i = 0; i < n; ++i) {
      scanf("%lld", &b[i]);
      s += b[i];
    }
    for (int i = 0; i < n; ++i) {
      scanf("%lld", &c[i]);
      s += c[i];
    }
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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 maxn = 200005;
int n;
long long b[maxn], c[maxn], s, a[maxn], num[63];
bool check() {
  memset(num, 0, sizeof num);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 63; ++j)
      if ((a[i] >> j) & 1) ++num[j];
  for (int i = 0; i < n; ++i) {
    long long B = 0;
    for (int j = 0; j < 63; ++j)
      if ((a[i] >> j) & 1) B += (1 << j) * num[j];
    if (B != b[i]) return false;
  }
  for (int i = 0; i < n; ++i) {
    long long C = 0;
    for (int j = 0; j < 63; ++j)
      if ((a[i] >> j) & 1)
        C += (1 << j) * n;
      else
        C += (1 << j) * num[j];
    if (C != c[i]) return false;
  }
  return true;
}
void solve() {
  if (s % (n << 1)) {
    puts("-1");
    return;
  }
  s /= (n << 1);
  for (int i = 0; i < n; ++i) {
    if (b[i] + c[i] - s < 0 || (b[i] + c[i] - s) % n) {
      puts("-1");
      return;
    }
    a[i] = (b[i] + c[i] - s) / n;
  }
  if (!check()) {
    puts("-1");
    return;
  }
  for (int i = 0; i < n; ++i) printf("%lld%c", a[i], i + 1 == n ? '\n' : ' ');
}
int main() {
  while (scanf("%d", &n) == 1) {
    s = 0;
    for (int i = 0; i < n; ++i) {
      scanf("%lld", &b[i]);
      s += b[i];
    }
    for (int i = 0; i < n; ++i) {
      scanf("%lld", &c[i]);
      s += c[i];
    }
    solve();
  }
  return 0;
}

```