#include <bits/stdc++.h>
using namespace std;
const int MAXN = 212345;
int b[MAXN], c[MAXN];
int a[MAXN];
int ones[32];
int main() {
  int n;
  int i, j;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &b[i]);
  for (i = 0; i < n; i++) scanf("%d", &c[i]);
  long long sum = 0;
  for (i = 0; i < n; i++) sum += b[i] + c[i];
  if (sum % (2 * n) != 0) {
    printf("-1\n");
    exit(0);
  }
  sum /= 2 * n;
  for (i = 0; i < n; i++) {
    if ((b[i] + c[i] - sum) % n != 0) {
      printf("-1\n");
      exit(0);
    }
    a[i] = (b[i] + c[i] - sum) / n;
    if (a[i] < 0 || (long long)a[i] * n != b[i] + c[i] - sum) {
      printf("-1\n");
      exit(0);
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < 31; j++) ones[j] += a[i] >> j & 1;
  }
  for (i = 0; i < n; i++) {
    long long t = 0;
    for (j = 0; j < 31; j++) {
      if (a[i] >> j & 1) {
        t += (1LL << j) * ones[j];
      }
    }
    if (t != b[i]) {
      printf("-1\n");
      exit(0);
    }
    t = 0;
    for (j = 0; j < 31; j++) {
      if (a[i] >> j & 1) {
        t += (1LL << j) * n;
      } else {
        t += (1LL << j) * ones[j];
      }
    }
    if (t != c[i]) {
      printf("-1\n");
      exit(0);
    }
  }
  for (i = 0; i < n; i++) printf("%d ", a[i]);
  puts("");
  return 0;
}

### Prompt

Please create a solution in CPP to 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 MAXN = 212345;
int b[MAXN], c[MAXN];
int a[MAXN];
int ones[32];
int main() {
  int n;
  int i, j;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &b[i]);
  for (i = 0; i < n; i++) scanf("%d", &c[i]);
  long long sum = 0;
  for (i = 0; i < n; i++) sum += b[i] + c[i];
  if (sum % (2 * n) != 0) {
    printf("-1\n");
    exit(0);
  }
  sum /= 2 * n;
  for (i = 0; i < n; i++) {
    if ((b[i] + c[i] - sum) % n != 0) {
      printf("-1\n");
      exit(0);
    }
    a[i] = (b[i] + c[i] - sum) / n;
    if (a[i] < 0 || (long long)a[i] * n != b[i] + c[i] - sum) {
      printf("-1\n");
      exit(0);
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < 31; j++) ones[j] += a[i] >> j & 1;
  }
  for (i = 0; i < n; i++) {
    long long t = 0;
    for (j = 0; j < 31; j++) {
      if (a[i] >> j & 1) {
        t += (1LL << j) * ones[j];
      }
    }
    if (t != b[i]) {
      printf("-1\n");
      exit(0);
    }
    t = 0;
    for (j = 0; j < 31; j++) {
      if (a[i] >> j & 1) {
        t += (1LL << j) * n;
      } else {
        t += (1LL << j) * ones[j];
      }
    }
    if (t != c[i]) {
      printf("-1\n");
      exit(0);
    }
  }
  for (i = 0; i < n; i++) printf("%d ", a[i]);
  puts("");
  return 0;
}

```