#include <bits/stdc++.h>
inline int get_ui();
int get_ui() {
  char c;
  for (; (c = getchar()) < '0' || c > '9';)
    ;
  for (int a = c ^ '0';; a = (a * 10) + (c ^ '0'))
    if ((c = getchar()) < '0' || c > '9') return a;
}
inline void put_ull(long long x);
void put_ull(long long x) {
  char buf[24], *p = buf;
  do *p++ = '0' ^ x % 10;
  while (x /= 10);
  do putchar(*--p);
  while (p > buf);
  putchar(' ');
}
int a[200000], b[200000], c[200000];
int bad() {
  puts("-1");
  return 0;
}
int main() {
  int n = get_ui();
  for (int i = 0; i < n; i++) b[i] = get_ui();
  long long lls = 0;
  for (int i = 0; i < n; i++) c[i] = get_ui(), lls += b[i] + c[i];
  if (lls % (2 * n)) return bad();
  int s = lls / (2 * n), mx = 0;
  for (int i = 0; i < n; i++) {
    int x = b[i] + c[i] - s;
    if (x % n) return bad();
    x /= n;
    a[i] = x;
    if (mx < x) mx = x;
  }
  for (int w = 1; w <= mx; w <<= 1) {
    int wn = w * n, s = 0;
    for (int i = 0; i < n; i++)
      if (a[i] & w) s++;
    s *= w;
    for (int i = 0; i < n; i++)
      if (a[i] & w) {
        b[i] -= s;
        c[i] -= wn;
      } else
        c[i] -= s;
  }
  for (int i = 0; i < n; i++)
    if (b[i] || c[i]) return bad();
  for (int i = 0; i < n; i++) put_ull(a[i]);
}

### Prompt

Please formulate 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>
inline int get_ui();
int get_ui() {
  char c;
  for (; (c = getchar()) < '0' || c > '9';)
    ;
  for (int a = c ^ '0';; a = (a * 10) + (c ^ '0'))
    if ((c = getchar()) < '0' || c > '9') return a;
}
inline void put_ull(long long x);
void put_ull(long long x) {
  char buf[24], *p = buf;
  do *p++ = '0' ^ x % 10;
  while (x /= 10);
  do putchar(*--p);
  while (p > buf);
  putchar(' ');
}
int a[200000], b[200000], c[200000];
int bad() {
  puts("-1");
  return 0;
}
int main() {
  int n = get_ui();
  for (int i = 0; i < n; i++) b[i] = get_ui();
  long long lls = 0;
  for (int i = 0; i < n; i++) c[i] = get_ui(), lls += b[i] + c[i];
  if (lls % (2 * n)) return bad();
  int s = lls / (2 * n), mx = 0;
  for (int i = 0; i < n; i++) {
    int x = b[i] + c[i] - s;
    if (x % n) return bad();
    x /= n;
    a[i] = x;
    if (mx < x) mx = x;
  }
  for (int w = 1; w <= mx; w <<= 1) {
    int wn = w * n, s = 0;
    for (int i = 0; i < n; i++)
      if (a[i] & w) s++;
    s *= w;
    for (int i = 0; i < n; i++)
      if (a[i] & w) {
        b[i] -= s;
        c[i] -= wn;
      } else
        c[i] -= s;
  }
  for (int i = 0; i < n; i++)
    if (b[i] || c[i]) return bad();
  for (int i = 0; i < n; i++) put_ull(a[i]);
}

```