#include <bits/stdc++.h>
using namespace std;
const int N = 200200;
int n;
long long b[N], c[N];
long long a[N];
int dg[32];
void solve() {
  memset((dg), (0), sizeof(dg));
  long long sum = 0, suma = 0;
  for (int i = (int)0; i < (int)n; ++i) sum += b[i] + c[i];
  suma = sum / (n + n);
  bool flag = 1;
  for (int i = (int)0; i < (int)n; ++i) {
    a[i] = (b[i] + c[i] - suma) / n;
    if (a[i] < 0) flag = 0;
  }
  if (!flag) {
    cout << "-1";
    return;
  }
  for (int i = (int)0; i < (int)n; ++i) {
    int t = 0;
    for (long long x = a[i]; x; x >>= 1) dg[t++] += (x & 1);
  }
  for (int i = (int)0; i < (int)n; ++i) {
    long long x = a[i], sumi = 0;
    for (int j = (int)0; j < (int)32; ++j)
      if ((x >> j) & 1) sumi += (1 << j) * dg[j];
    if (sumi != b[i]) {
      flag = 0;
      break;
    }
  }
  if (!flag) {
    cout << "-1";
    return;
  }
  for (int i = (int)0; i < (int)n; ++i) {
    long long x = a[i], sumi = 0;
    for (int j = (int)0; j < (int)32; ++j) {
      if ((x >> j) & 1)
        sumi += (1 << j) * n;
      else
        sumi += (1 << j) * dg[j];
    }
    if (sumi != c[i]) {
      flag = 0;
      break;
    }
  }
  if (!flag) {
    cout << "-1";
    return;
  }
  for (int i = (int)0; i < (int)n; ++i) cout << a[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = (int)0; i < (int)n; ++i) cin >> b[i];
  for (int i = (int)0; i < (int)n; ++i) cin >> c[i];
  solve();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 200200;
int n;
long long b[N], c[N];
long long a[N];
int dg[32];
void solve() {
  memset((dg), (0), sizeof(dg));
  long long sum = 0, suma = 0;
  for (int i = (int)0; i < (int)n; ++i) sum += b[i] + c[i];
  suma = sum / (n + n);
  bool flag = 1;
  for (int i = (int)0; i < (int)n; ++i) {
    a[i] = (b[i] + c[i] - suma) / n;
    if (a[i] < 0) flag = 0;
  }
  if (!flag) {
    cout << "-1";
    return;
  }
  for (int i = (int)0; i < (int)n; ++i) {
    int t = 0;
    for (long long x = a[i]; x; x >>= 1) dg[t++] += (x & 1);
  }
  for (int i = (int)0; i < (int)n; ++i) {
    long long x = a[i], sumi = 0;
    for (int j = (int)0; j < (int)32; ++j)
      if ((x >> j) & 1) sumi += (1 << j) * dg[j];
    if (sumi != b[i]) {
      flag = 0;
      break;
    }
  }
  if (!flag) {
    cout << "-1";
    return;
  }
  for (int i = (int)0; i < (int)n; ++i) {
    long long x = a[i], sumi = 0;
    for (int j = (int)0; j < (int)32; ++j) {
      if ((x >> j) & 1)
        sumi += (1 << j) * n;
      else
        sumi += (1 << j) * dg[j];
    }
    if (sumi != c[i]) {
      flag = 0;
      break;
    }
  }
  if (!flag) {
    cout << "-1";
    return;
  }
  for (int i = (int)0; i < (int)n; ++i) cout << a[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = (int)0; i < (int)n; ++i) cin >> b[i];
  for (int i = (int)0; i < (int)n; ++i) cin >> c[i];
  solve();
  return 0;
}

```