#include <bits/stdc++.h>
using namespace std;
int b[201000], c[201000];
long long a[201000], pow2[70];
int cnt[70], num[70];
int main() {
  pow2[0] = 1;
  for (int i = 1; i < 70; i++) pow2[i] = pow2[i - 1] << 1;
  int n;
  scanf("%d", &n);
  long long sum = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]), sum += b[i];
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]), sum += c[i];
  if (sum % (2 * n)) {
    puts("-1");
    return 0;
  }
  long long suma = sum / (2 * n);
  for (int i = 1; i <= n; i++) {
    long long tmp = (long long)b[i] + c[i] - suma;
    if (tmp % n || tmp < 0) {
      puts("-1");
      return 0;
    }
    a[i] = tmp / n;
  }
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= n; i++) {
    long long x = a[i];
    int len = 0;
    while (x) cnt[len++] += x & 1, x >>= 1;
  }
  for (int i = 1; i <= n; i++) {
    long long x = a[i];
    int len = 0;
    memset(num, 0, sizeof(num));
    while (x) num[len++] += x & 1, x >>= 1;
    long long ans = 0;
    for (int j = 0; j < 70; j++) {
      int t = cnt[j];
      if (num[j]) ans += pow2[j] * t;
    }
    if (ans != b[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long x = a[i];
    int len = 0;
    memset(num, 0, sizeof(num));
    while (x) num[len++] += x & 1, x >>= 1;
    long long ans = 0;
    for (int j = 0; j < 70; j++) {
      int t = num[j] ? n : cnt[j];
      ans += pow2[j] * t;
    }
    if (ans != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
int b[201000], c[201000];
long long a[201000], pow2[70];
int cnt[70], num[70];
int main() {
  pow2[0] = 1;
  for (int i = 1; i < 70; i++) pow2[i] = pow2[i - 1] << 1;
  int n;
  scanf("%d", &n);
  long long sum = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]), sum += b[i];
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]), sum += c[i];
  if (sum % (2 * n)) {
    puts("-1");
    return 0;
  }
  long long suma = sum / (2 * n);
  for (int i = 1; i <= n; i++) {
    long long tmp = (long long)b[i] + c[i] - suma;
    if (tmp % n || tmp < 0) {
      puts("-1");
      return 0;
    }
    a[i] = tmp / n;
  }
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= n; i++) {
    long long x = a[i];
    int len = 0;
    while (x) cnt[len++] += x & 1, x >>= 1;
  }
  for (int i = 1; i <= n; i++) {
    long long x = a[i];
    int len = 0;
    memset(num, 0, sizeof(num));
    while (x) num[len++] += x & 1, x >>= 1;
    long long ans = 0;
    for (int j = 0; j < 70; j++) {
      int t = cnt[j];
      if (num[j]) ans += pow2[j] * t;
    }
    if (ans != b[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long x = a[i];
    int len = 0;
    memset(num, 0, sizeof(num));
    while (x) num[len++] += x & 1, x >>= 1;
    long long ans = 0;
    for (int j = 0; j < 70; j++) {
      int t = num[j] ? n : cnt[j];
      ans += pow2[j] * t;
    }
    if (ans != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}

```