#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200010;
int n;
int cnt[70];
long long a[MAX_N], b[MAX_N], c[MAX_N];
bool solve() {
  if (n == 1) {
    if ((b[1] | b[1]) == c[1]) {
      a[1] = b[1];
      return true;
    } else
      return false;
  }
  long long sum = 0;
  for (int i = 1; i <= n; ++i) {
    sum += b[i] + c[i];
  }
  if (sum % (2 * n)) return false;
  sum /= (2 * n);
  for (int i = 1; i <= n; ++i) {
    long long left = b[i] + c[i] - sum;
    if (left < 0 || left % n) return false;
    a[i] = left / n;
  }
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 60; ++j) {
      if ((a[i] >> j) & 1) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; ++i) {
    long long tb = 0, tc = 0;
    for (int j = 0; j <= 60; ++j) {
      if ((a[i] >> j) & 1) {
        tb += 1ll * cnt[j] * (1ll << j);
        tc += 1ll * n * (1ll << j);
      } else
        tc += 1ll * cnt[j] * (1ll << j);
    }
    if (tb != b[i] || tc != c[i]) return false;
  }
  return true;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; ++i) scanf("%lld", &b[i]);
    for (int i = 1; i <= n; ++i) scanf("%lld", &c[i]);
    if (solve() == false)
      printf("-1");
    else
      for (int i = 1; i <= n; ++i) printf("%lld ", a[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
const int MAX_N = 200010;
int n;
int cnt[70];
long long a[MAX_N], b[MAX_N], c[MAX_N];
bool solve() {
  if (n == 1) {
    if ((b[1] | b[1]) == c[1]) {
      a[1] = b[1];
      return true;
    } else
      return false;
  }
  long long sum = 0;
  for (int i = 1; i <= n; ++i) {
    sum += b[i] + c[i];
  }
  if (sum % (2 * n)) return false;
  sum /= (2 * n);
  for (int i = 1; i <= n; ++i) {
    long long left = b[i] + c[i] - sum;
    if (left < 0 || left % n) return false;
    a[i] = left / n;
  }
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 60; ++j) {
      if ((a[i] >> j) & 1) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; ++i) {
    long long tb = 0, tc = 0;
    for (int j = 0; j <= 60; ++j) {
      if ((a[i] >> j) & 1) {
        tb += 1ll * cnt[j] * (1ll << j);
        tc += 1ll * n * (1ll << j);
      } else
        tc += 1ll * cnt[j] * (1ll << j);
    }
    if (tb != b[i] || tc != c[i]) return false;
  }
  return true;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; ++i) scanf("%lld", &b[i]);
    for (int i = 1; i <= n; ++i) scanf("%lld", &c[i]);
    if (solve() == false)
      printf("-1");
    else
      for (int i = 1; i <= n; ++i) printf("%lld ", a[i]);
    printf("\n");
  }
  return 0;
}

```