#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 88;
const int logn = 40;
long long getll() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
void printll(long long x) { printf("%I64d ", x); }
long long a[maxn], b[maxn], c[maxn], d[maxn];
long long cnt[logn];
int main() {
  long long n, i, j, sum = 0, cur;
  n = getll();
  for (i = 0; i < n; i++) b[i] = getll();
  for (i = 0; i < n; i++) c[i] = getll();
  for (i = 0; i < n; i++) d[i] = b[i] + c[i];
  for (i = 0; i < n; i++) sum += d[i];
  sum /= 2 * n;
  for (i = 0; i < n; i++) {
    a[i] = (d[i] - sum) / n;
    if (a[i] < 0) {
      printll(-1);
      return 0;
    }
    cur = a[i];
    j = 0;
    while (cur) {
      if (cur & 1ll) cnt[j]++;
      cur >>= 1ll;
      j++;
    }
  }
  for (i = 0; i < n; i++) {
    long long curb = 0, curc = 0;
    for (j = 0; j < logn; j++)
      if ((a[i] >> j) & 1ll) {
        curb += (1ll << j) * cnt[j];
        curc += (1ll << j) * n;
      } else
        curc += (1ll << j) * cnt[j];
    if (curb != b[i] || curc != c[i]) {
      printll(-1);
      return 0;
    }
  }
  for (i = 0; i < n; i++) printll(a[i]);
  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 = 2e5 + 88;
const int logn = 40;
long long getll() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
void printll(long long x) { printf("%I64d ", x); }
long long a[maxn], b[maxn], c[maxn], d[maxn];
long long cnt[logn];
int main() {
  long long n, i, j, sum = 0, cur;
  n = getll();
  for (i = 0; i < n; i++) b[i] = getll();
  for (i = 0; i < n; i++) c[i] = getll();
  for (i = 0; i < n; i++) d[i] = b[i] + c[i];
  for (i = 0; i < n; i++) sum += d[i];
  sum /= 2 * n;
  for (i = 0; i < n; i++) {
    a[i] = (d[i] - sum) / n;
    if (a[i] < 0) {
      printll(-1);
      return 0;
    }
    cur = a[i];
    j = 0;
    while (cur) {
      if (cur & 1ll) cnt[j]++;
      cur >>= 1ll;
      j++;
    }
  }
  for (i = 0; i < n; i++) {
    long long curb = 0, curc = 0;
    for (j = 0; j < logn; j++)
      if ((a[i] >> j) & 1ll) {
        curb += (1ll << j) * cnt[j];
        curc += (1ll << j) * n;
      } else
        curc += (1ll << j) * cnt[j];
    if (curb != b[i] || curc != c[i]) {
      printll(-1);
      return 0;
    }
  }
  for (i = 0; i < n; i++) printll(a[i]);
  return 0;
}

```