#include <bits/stdc++.h>
using namespace std;
const int dx[9] = {0, 1, -1, 0, 0, -1, -1, 1, 1};
const int dy[9] = {0, 0, 0, -1, 1, -1, 1, -1, 1};
const double pi = acos(-1.0);
const int N = 2e5 + 100;
int n;
long long b[N], c[N], d[N], sumd = 0, a[N], ans[N];
long long B[N][65], C[N][65];
long long bin[65];
void out() {
  cout << -1 << endl;
  exit(0);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  bin[0] = 1;
  for (int i = 1; i <= 62; i++) bin[i] = bin[i - 1] << 1;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    d[i] = b[i] + c[i];
    sumd += d[i];
  }
  for (int i = 1; i <= n; i++) {
    a[i] = 1LL * 2 * n * d[i] - sumd;
    long long temp = 1LL * 2 * n * n;
    if (a[i] % temp != 0) out();
    a[i] /= temp;
    if (a[i] < 0) out();
    ans[i] = a[i];
  }
  long long ma = *max_element(a + 1, a + 1 + n);
  int limit = 0;
  while (ma) {
    ma >>= 1;
    limit++;
  }
  for (int i = 0; i <= limit - 1; i++) {
    int k = 0;
    for (int j = 1; j <= n; j++)
      if (a[j] & bin[i]) k++;
    for (int j = 1; j <= n; j++) {
      if (a[j] & bin[i]) {
        B[j][i] += k, C[j][i] += n;
      } else
        C[j][i] += k;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long tb = 0;
    for (int j = 0; j <= limit - 1; j++) tb = tb + B[i][j] * bin[j];
    long long tc = 0;
    for (int j = 0; j <= limit - 1; j++) tc = tc + C[i][j] * bin[j];
    if (tb != b[i] || tc != c[i]) out();
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}

### Prompt

Create a solution in cpp for 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 dx[9] = {0, 1, -1, 0, 0, -1, -1, 1, 1};
const int dy[9] = {0, 0, 0, -1, 1, -1, 1, -1, 1};
const double pi = acos(-1.0);
const int N = 2e5 + 100;
int n;
long long b[N], c[N], d[N], sumd = 0, a[N], ans[N];
long long B[N][65], C[N][65];
long long bin[65];
void out() {
  cout << -1 << endl;
  exit(0);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  bin[0] = 1;
  for (int i = 1; i <= 62; i++) bin[i] = bin[i - 1] << 1;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    d[i] = b[i] + c[i];
    sumd += d[i];
  }
  for (int i = 1; i <= n; i++) {
    a[i] = 1LL * 2 * n * d[i] - sumd;
    long long temp = 1LL * 2 * n * n;
    if (a[i] % temp != 0) out();
    a[i] /= temp;
    if (a[i] < 0) out();
    ans[i] = a[i];
  }
  long long ma = *max_element(a + 1, a + 1 + n);
  int limit = 0;
  while (ma) {
    ma >>= 1;
    limit++;
  }
  for (int i = 0; i <= limit - 1; i++) {
    int k = 0;
    for (int j = 1; j <= n; j++)
      if (a[j] & bin[i]) k++;
    for (int j = 1; j <= n; j++) {
      if (a[j] & bin[i]) {
        B[j][i] += k, C[j][i] += n;
      } else
        C[j][i] += k;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long tb = 0;
    for (int j = 0; j <= limit - 1; j++) tb = tb + B[i][j] * bin[j];
    long long tc = 0;
    for (int j = 0; j <= limit - 1; j++) tc = tc + C[i][j] * bin[j];
    if (tb != b[i] || tc != c[i]) out();
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}

```