#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
const long double PI = 3.14159265;
long long int powerWithMod(long long int base, long long int exponent,
                           long long int modulus = LLONG_MAX) {
  long long int result = 1;
  base %= modulus;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent >>= 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long int modInverse(long long int a, long long int m = MOD) {
  return powerWithMod(a, m - 2, m);
}
long long int n, b[212345], c[212345], a[212345], cnt[32];
void t() {
  cout << "-1\n";
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    if (c[i] < b[i]) t();
  }
  auto s = accumulate(b, b + n, 0LL) + accumulate(c, c + n, 0LL);
  s /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = (b[i] + c[i] - s) / n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 32; j++)
      if (a[i] & (1LL << j)) cnt[j]++;
  for (int i = 0; i < n; i++) {
    long long int currb = 0, currc = 0;
    for (int j = 0; j < 32; j++) {
      if (a[i] & (1LL << j))
        currb += (1LL << j) * cnt[j], currc += (1LL << j) * n;
      else
        currc += (1LL << j) * cnt[j];
    }
    if (currb != b[i] or currc != c[i]) t();
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
  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 long long int MOD = 1e9 + 7;
const long double PI = 3.14159265;
long long int powerWithMod(long long int base, long long int exponent,
                           long long int modulus = LLONG_MAX) {
  long long int result = 1;
  base %= modulus;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent >>= 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long int modInverse(long long int a, long long int m = MOD) {
  return powerWithMod(a, m - 2, m);
}
long long int n, b[212345], c[212345], a[212345], cnt[32];
void t() {
  cout << "-1\n";
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    if (c[i] < b[i]) t();
  }
  auto s = accumulate(b, b + n, 0LL) + accumulate(c, c + n, 0LL);
  s /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = (b[i] + c[i] - s) / n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 32; j++)
      if (a[i] & (1LL << j)) cnt[j]++;
  for (int i = 0; i < n; i++) {
    long long int currb = 0, currc = 0;
    for (int j = 0; j < 32; j++) {
      if (a[i] & (1LL << j))
        currb += (1LL << j) * cnt[j], currc += (1LL << j) * n;
      else
        currc += (1LL << j) * cnt[j];
    }
    if (currb != b[i] or currc != c[i]) t();
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
  return 0;
}

```