#include <bits/stdc++.h>
using namespace std;
const int S = 200003;
char buf[1000000], *p1, *p2;
inline char gc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int rd() {
  register int f = 0;
  register char c = gc();
  while (c < 48 || c > 57) c = gc();
  while (c > 47 && c < 58) f = f * 10 + (c ^ 48), c = gc();
  return f;
}
int a[S], b[S], c[S];
int main() {
  register int n = rd(), i, j, d;
  long long s = 0;
  for (i = 1; i <= n; ++i) s += (b[i] = rd());
  for (i = 1; i <= n; ++i) s += (c[i] = rd());
  s /= (n << 1);
  for (i = 1; i <= n; ++i) a[i] = (b[i] + c[i] - s) / n;
  for (i = 0; i < 31; ++i) {
    d = 0;
    for (j = 1; j <= n; ++j)
      if (a[j] & (1 << i)) ++d, c[j] -= n * (1 << i);
    for (j = 1; j <= n; ++j)
      if (a[j] & (1 << i))
        b[j] -= d * (1 << i);
      else
        c[j] -= d * (1 << i);
  }
  for (i = 1; i <= n; ++i)
    if (b[i] || c[i]) {
      printf("-1");
      return 0;
    }
  for (i = 1; i <= n; ++i) printf("%d ", a[i]);
  return 0;
}

### Prompt

Generate 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 S = 200003;
char buf[1000000], *p1, *p2;
inline char gc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int rd() {
  register int f = 0;
  register char c = gc();
  while (c < 48 || c > 57) c = gc();
  while (c > 47 && c < 58) f = f * 10 + (c ^ 48), c = gc();
  return f;
}
int a[S], b[S], c[S];
int main() {
  register int n = rd(), i, j, d;
  long long s = 0;
  for (i = 1; i <= n; ++i) s += (b[i] = rd());
  for (i = 1; i <= n; ++i) s += (c[i] = rd());
  s /= (n << 1);
  for (i = 1; i <= n; ++i) a[i] = (b[i] + c[i] - s) / n;
  for (i = 0; i < 31; ++i) {
    d = 0;
    for (j = 1; j <= n; ++j)
      if (a[j] & (1 << i)) ++d, c[j] -= n * (1 << i);
    for (j = 1; j <= n; ++j)
      if (a[j] & (1 << i))
        b[j] -= d * (1 << i);
      else
        c[j] -= d * (1 << i);
  }
  for (i = 1; i <= n; ++i)
    if (b[i] || c[i]) {
      printf("-1");
      return 0;
    }
  for (i = 1; i <= n; ++i) printf("%d ", a[i]);
  return 0;
}

```