#include <bits/stdc++.h>
long long a[200005], b[200005], c[200005], cnt[105];
int main() {
  long long n;
  while (scanf("%I64d", &n) != EOF) {
    long long i, j, sum = 0;
    for (i = 0; i < n; i++) {
      scanf("%I64d", &b[i]);
      sum += b[i];
    }
    for (i = 0; i < n; i++) {
      scanf("%I64d", &c[i]);
      sum += c[i];
    }
    long long zt = 0;
    if (sum % (2 * n))
      zt = 1;
    else
      sum /= (2 * n);
    for (i = 0; i < n; i++) {
      if (zt || (b[i] + c[i] - sum) % n || b[i] + c[i] < sum) {
        zt = 1;
        break;
      }
      a[i] = (b[i] + c[i] - sum) / n;
    }
    memset(cnt, 0, sizeof(cnt));
    for (i = 0; i < n; i++) {
      for (j = 0; j < 60; j++) {
        if ((a[i] >> j) & 1) cnt[j]++;
      }
    }
    long long sumb, sumc;
    for (i = 0; i < n; i++) {
      sumb = 0;
      sumc = 0;
      for (j = 0; j < 60; j++) {
        if ((a[i] >> j) & 1) {
          sumb += cnt[j] * (1LL << j);
          sumc += n * (1LL << j);
        } else
          sumc += cnt[j] * (1LL << j);
      }
      if (zt || sumb != b[i] || sumc != c[i]) {
        zt = 1;
        break;
      }
    }
    if (zt)
      printf("-1\n");
    else {
      for (i = 0; i < n; i++) {
        if (i) printf(" ");
        printf("%I64d", a[i]);
      }
      printf("\n");
    }
  }
}

### Prompt

Please formulate a cpp solution 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>
long long a[200005], b[200005], c[200005], cnt[105];
int main() {
  long long n;
  while (scanf("%I64d", &n) != EOF) {
    long long i, j, sum = 0;
    for (i = 0; i < n; i++) {
      scanf("%I64d", &b[i]);
      sum += b[i];
    }
    for (i = 0; i < n; i++) {
      scanf("%I64d", &c[i]);
      sum += c[i];
    }
    long long zt = 0;
    if (sum % (2 * n))
      zt = 1;
    else
      sum /= (2 * n);
    for (i = 0; i < n; i++) {
      if (zt || (b[i] + c[i] - sum) % n || b[i] + c[i] < sum) {
        zt = 1;
        break;
      }
      a[i] = (b[i] + c[i] - sum) / n;
    }
    memset(cnt, 0, sizeof(cnt));
    for (i = 0; i < n; i++) {
      for (j = 0; j < 60; j++) {
        if ((a[i] >> j) & 1) cnt[j]++;
      }
    }
    long long sumb, sumc;
    for (i = 0; i < n; i++) {
      sumb = 0;
      sumc = 0;
      for (j = 0; j < 60; j++) {
        if ((a[i] >> j) & 1) {
          sumb += cnt[j] * (1LL << j);
          sumc += n * (1LL << j);
        } else
          sumc += cnt[j] * (1LL << j);
      }
      if (zt || sumb != b[i] || sumc != c[i]) {
        zt = 1;
        break;
      }
    }
    if (zt)
      printf("-1\n");
    else {
      for (i = 0; i < n; i++) {
        if (i) printf(" ");
        printf("%I64d", a[i]);
      }
      printf("\n");
    }
  }
}

```