#include <bits/stdc++.h>
using namespace std;
int n, b[200011], c[200011], d[200011], cnt[200011], temp, step, bb, cc, maxx;
long long sn;
bool flag;
int main(void) {
  sn = 0;
  flag = false;
  maxx = -1;
  memset(cnt, 0, sizeof(cnt));
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    sn += b[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    d[i] = b[i] + c[i];
    sn += c[i];
  }
  if (sn % (2 * n)) {
    cout << "-1" << endl;
    return 0;
  }
  sn = sn / (2 * n);
  for (int i = 1; i <= n; i++) {
    d[i] = d[i] - sn;
    if (d[i] % n) {
      cout << "-1" << endl;
      return 0;
    }
    d[i] = d[i] / n;
    if (d[i] < 0) {
      cout << "-1" << endl;
      return 0;
    }
    temp = d[i];
    step = 0;
    while (temp) {
      if (temp & 1) {
        cnt[step]++;
      }
      step++;
      temp = temp >> 1;
    }
    maxx = max(maxx, step - 1);
  }
  for (int i = 1; i <= n; i++) {
    temp = d[i];
    step = 0;
    bb = 0;
    cc = 0;
    while (temp || step <= maxx) {
      if (temp & 1) {
        bb = bb + cnt[step] * (1 << step);
        cc = cc + n * (1 << step);
      } else {
        bb = bb + 0;
        cc = cc + cnt[step] * (1 << step);
      }
      step++;
      temp = temp >> 1;
    }
    if (bb != b[i] || cc != c[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << d[i] << " ";
  }
  cout << endl;
  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;
int n, b[200011], c[200011], d[200011], cnt[200011], temp, step, bb, cc, maxx;
long long sn;
bool flag;
int main(void) {
  sn = 0;
  flag = false;
  maxx = -1;
  memset(cnt, 0, sizeof(cnt));
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    sn += b[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    d[i] = b[i] + c[i];
    sn += c[i];
  }
  if (sn % (2 * n)) {
    cout << "-1" << endl;
    return 0;
  }
  sn = sn / (2 * n);
  for (int i = 1; i <= n; i++) {
    d[i] = d[i] - sn;
    if (d[i] % n) {
      cout << "-1" << endl;
      return 0;
    }
    d[i] = d[i] / n;
    if (d[i] < 0) {
      cout << "-1" << endl;
      return 0;
    }
    temp = d[i];
    step = 0;
    while (temp) {
      if (temp & 1) {
        cnt[step]++;
      }
      step++;
      temp = temp >> 1;
    }
    maxx = max(maxx, step - 1);
  }
  for (int i = 1; i <= n; i++) {
    temp = d[i];
    step = 0;
    bb = 0;
    cc = 0;
    while (temp || step <= maxx) {
      if (temp & 1) {
        bb = bb + cnt[step] * (1 << step);
        cc = cc + n * (1 << step);
      } else {
        bb = bb + 0;
        cc = cc + cnt[step] * (1 << step);
      }
      step++;
      temp = temp >> 1;
    }
    if (bb != b[i] || cc != c[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << d[i] << " ";
  }
  cout << endl;
  return 0;
}

```