#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int f = 1, r = 0;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) r = (r << 1) + (r << 3) + (ch ^ 48), ch = getchar();
  return f ? r : -r;
}
const int N = 2e5 + 5;
int n, nn;
long long cnt[40], pw[40], a[N], b[N], c[N];
long long sum;
int main() {
  n = read(), nn = n << 1;
  for (int i = 1; i <= n; i++) sum += b[i] = read();
  for (int i = 1; i <= n; i++) sum += c[i] = read();
  if (sum % nn) puts("-1"), exit(0);
  sum /= nn;
  for (int i = 1; i <= n; i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] % n) puts("-1"), exit(0);
    a[i] /= n;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 30; ~j; j--)
      if (a[i] & (1 << j)) cnt[j]++;
  for (int i = pw[0] = 1; i <= 30; i++) pw[i] = pw[i - 1] << 1;
  for (int i = 1; i <= n; i++) {
    long long res = 0;
    for (int j = 30; ~j; j--)
      if (a[i] & (1 << j)) {
        res += 1ll * pw[j] * cnt[j];
        if (res > b[i]) puts("-1"), exit(0);
      }
    if (res ^ b[i]) puts("-1"), exit(0);
    res = sum;
    for (int j = 30; ~j; j--)
      if (a[i] & (1 << j)) {
        res += 1ll * pw[j] * (n - cnt[j]);
        if (res > c[i]) puts("-1"), exit(0);
      }
    if (res ^ c[i]) puts("-1"), exit(0);
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
  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;
inline int read() {
  int f = 1, r = 0;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) r = (r << 1) + (r << 3) + (ch ^ 48), ch = getchar();
  return f ? r : -r;
}
const int N = 2e5 + 5;
int n, nn;
long long cnt[40], pw[40], a[N], b[N], c[N];
long long sum;
int main() {
  n = read(), nn = n << 1;
  for (int i = 1; i <= n; i++) sum += b[i] = read();
  for (int i = 1; i <= n; i++) sum += c[i] = read();
  if (sum % nn) puts("-1"), exit(0);
  sum /= nn;
  for (int i = 1; i <= n; i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] % n) puts("-1"), exit(0);
    a[i] /= n;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 30; ~j; j--)
      if (a[i] & (1 << j)) cnt[j]++;
  for (int i = pw[0] = 1; i <= 30; i++) pw[i] = pw[i - 1] << 1;
  for (int i = 1; i <= n; i++) {
    long long res = 0;
    for (int j = 30; ~j; j--)
      if (a[i] & (1 << j)) {
        res += 1ll * pw[j] * cnt[j];
        if (res > b[i]) puts("-1"), exit(0);
      }
    if (res ^ b[i]) puts("-1"), exit(0);
    res = sum;
    for (int j = 30; ~j; j--)
      if (a[i] & (1 << j)) {
        res += 1ll * pw[j] * (n - cnt[j]);
        if (res > c[i]) puts("-1"), exit(0);
      }
    if (res ^ c[i]) puts("-1"), exit(0);
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
  return 0;
}

```