#include <bits/stdc++.h>
using namespace std;
const long long INF = 9e18;
const long long MAX = 2e5 + 5;
const long long MOD = 1e9 + 7;
long long n;
long long a[MAX], b[MAX], c[MAX], bc[MAX];
long long bit[30];
long long sum;
void read() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    bc[i] = c[i];
    bc[i] += b[i];
  }
}
void build() {
  for (int i = 0; i < n; i++) sum += bc[i];
  sum /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = (bc[i] - sum) / n;
}
bool cb(int k) {
  int t = b[k];
  int now = 0;
  for (int i = 0; i < 30; i++) {
    if (a[k] & (1 << i)) now += bit[i];
    if (((t >> i) & 1) != now % 2) return 0;
    now /= 2;
  }
  return 1;
}
bool cc(int k) {
  int t = c[k];
  int now = 0;
  for (int i = 0; i < 30; i++) {
    if (a[k] & (1 << i))
      now += n;
    else
      now += bit[i];
    if (((t >> i) & 1) != now % 2) return 0;
    now /= 2;
  }
  return 1;
}
bool che() {
  for (int i = 0; i < n; i++)
    if (a[i] > 1000000000 || a[i] < 0) return 0;
  for (int i = 0; i < 30; i++) {
    for (int j = 0; j < n; j++) {
      if (a[j] & (1 << i)) bit[i]++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (!cb(i)) return 0;
  }
  for (int i = 0; i < n; i++) {
    if (!cc(i)) return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read();
  build();
  if (che()) {
    for (int i = 0; i < n; i++) cout << a[i] << ' ';
    cout << '\n';
  } else
    cout << "-1\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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 INF = 9e18;
const long long MAX = 2e5 + 5;
const long long MOD = 1e9 + 7;
long long n;
long long a[MAX], b[MAX], c[MAX], bc[MAX];
long long bit[30];
long long sum;
void read() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    bc[i] = c[i];
    bc[i] += b[i];
  }
}
void build() {
  for (int i = 0; i < n; i++) sum += bc[i];
  sum /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = (bc[i] - sum) / n;
}
bool cb(int k) {
  int t = b[k];
  int now = 0;
  for (int i = 0; i < 30; i++) {
    if (a[k] & (1 << i)) now += bit[i];
    if (((t >> i) & 1) != now % 2) return 0;
    now /= 2;
  }
  return 1;
}
bool cc(int k) {
  int t = c[k];
  int now = 0;
  for (int i = 0; i < 30; i++) {
    if (a[k] & (1 << i))
      now += n;
    else
      now += bit[i];
    if (((t >> i) & 1) != now % 2) return 0;
    now /= 2;
  }
  return 1;
}
bool che() {
  for (int i = 0; i < n; i++)
    if (a[i] > 1000000000 || a[i] < 0) return 0;
  for (int i = 0; i < 30; i++) {
    for (int j = 0; j < n; j++) {
      if (a[j] & (1 << i)) bit[i]++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (!cb(i)) return 0;
  }
  for (int i = 0; i < n; i++) {
    if (!cc(i)) return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read();
  build();
  if (che()) {
    for (int i = 0; i < n; i++) cout << a[i] << ' ';
    cout << '\n';
  } else
    cout << "-1\n";
  return 0;
}

```