#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 200005;
long long n, a[MAXN], b[MAXN], c[MAXN], bitsOfA[35];
void buildBitsArray() {
  for (long long(i) = (long long)(0); (i) <= (long long)(32); (i)++)
    for (long long(j) = (long long)(1); (j) <= (long long)(n); (j)++)
      bitsOfA[i] += ((a[j] & (1ll << i)) > 0);
}
bool check() {
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++) {
    long long curC = 0, curB = 0;
    for (long long(j) = (long long)(0); (j) <= (long long)(32); (j)++)
      curC += (n * (a[i] & (1ll << j)) + bitsOfA[j] * (~a[i] & (1ll << j))),
          curB += bitsOfA[j] * (a[i] & (1ll << j));
    if (curC != c[i] || curB != b[i]) return false;
  }
  return true;
}
int main() {
  scanf("%lld", &n);
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++)
    scanf("%lld", &b[i]);
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++)
    scanf("%lld", &c[i]);
  long long sumD = 0;
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++)
    a[i] = b[i] + c[i], sumD += a[i];
  if (sumD % (2 * n)) {
    printf("-1\n");
    return 0;
  }
  long long sumA = sumD / n / 2;
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++) {
    a[i] -= sumA;
    if (a[i] < 0 || a[i] % n) {
      printf("-1\n");
      return 0;
    }
    a[i] /= n;
  }
  buildBitsArray();
  if (check())
    for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++)
      printf("%lld ", a[i]);
  else
    printf("-1");
  printf("\n");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long MAXN = 200005;
long long n, a[MAXN], b[MAXN], c[MAXN], bitsOfA[35];
void buildBitsArray() {
  for (long long(i) = (long long)(0); (i) <= (long long)(32); (i)++)
    for (long long(j) = (long long)(1); (j) <= (long long)(n); (j)++)
      bitsOfA[i] += ((a[j] & (1ll << i)) > 0);
}
bool check() {
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++) {
    long long curC = 0, curB = 0;
    for (long long(j) = (long long)(0); (j) <= (long long)(32); (j)++)
      curC += (n * (a[i] & (1ll << j)) + bitsOfA[j] * (~a[i] & (1ll << j))),
          curB += bitsOfA[j] * (a[i] & (1ll << j));
    if (curC != c[i] || curB != b[i]) return false;
  }
  return true;
}
int main() {
  scanf("%lld", &n);
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++)
    scanf("%lld", &b[i]);
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++)
    scanf("%lld", &c[i]);
  long long sumD = 0;
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++)
    a[i] = b[i] + c[i], sumD += a[i];
  if (sumD % (2 * n)) {
    printf("-1\n");
    return 0;
  }
  long long sumA = sumD / n / 2;
  for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++) {
    a[i] -= sumA;
    if (a[i] < 0 || a[i] % n) {
      printf("-1\n");
      return 0;
    }
    a[i] /= n;
  }
  buildBitsArray();
  if (check())
    for (long long(i) = (long long)(1); (i) <= (long long)(n); (i)++)
      printf("%lld ", a[i]);
  else
    printf("-1");
  printf("\n");
  return 0;
}

```