#include <bits/stdc++.h>
using namespace std;
const long long N = 200001;
long long a, all, x, y, s1[N], s2[N], s[N], p[33];
inline long long r() {
  char ch = getchar();
  long long f1 = 0, f2 = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f2 = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    f1 = (f1 << 3) + (f1 << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f1 * f2;
}
signed main() {
  a = r();
  for (long long i = 1; i <= a; i++) s1[i] = r();
  for (long long i = 1; i <= a; i++) {
    s2[i] = r();
    if (s2[i] < s1[i]) {
      puts("-1");
      return 0;
    }
    all += s1[i] + s2[i];
  }
  if (all % (2 * a)) {
    puts("-1");
    return 0;
  }
  all /= 2 * a;
  for (long long i = 1; i <= a; i++) {
    s[i] = s1[i] + s2[i] - all;
    if (s[i] % a) {
      puts("-1");
      return 0;
    } else
      s[i] /= a;
  }
  for (long long i = 1; i <= a; i++)
    for (long long j = 1, k = s[i]; k; k >>= 1, j++)
      if (k & 1) p[j]++;
  for (long long i = 1; i <= a; i++) {
    x = y = 0;
    for (long long j = 1, k = s[i]; j <= 32; k >>= 1, j++)
      if (k & 1)
        x += p[j] << j - 1, y += a << j - 1;
      else
        y += p[j] << j - 1;
    if (x != s1[i] || y != s2[i]) {
      puts("-1");
      return 0;
    }
  }
  for (long long i = 1; i <= a; i++) printf("%lld ", s[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 long long N = 200001;
long long a, all, x, y, s1[N], s2[N], s[N], p[33];
inline long long r() {
  char ch = getchar();
  long long f1 = 0, f2 = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f2 = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    f1 = (f1 << 3) + (f1 << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f1 * f2;
}
signed main() {
  a = r();
  for (long long i = 1; i <= a; i++) s1[i] = r();
  for (long long i = 1; i <= a; i++) {
    s2[i] = r();
    if (s2[i] < s1[i]) {
      puts("-1");
      return 0;
    }
    all += s1[i] + s2[i];
  }
  if (all % (2 * a)) {
    puts("-1");
    return 0;
  }
  all /= 2 * a;
  for (long long i = 1; i <= a; i++) {
    s[i] = s1[i] + s2[i] - all;
    if (s[i] % a) {
      puts("-1");
      return 0;
    } else
      s[i] /= a;
  }
  for (long long i = 1; i <= a; i++)
    for (long long j = 1, k = s[i]; k; k >>= 1, j++)
      if (k & 1) p[j]++;
  for (long long i = 1; i <= a; i++) {
    x = y = 0;
    for (long long j = 1, k = s[i]; j <= 32; k >>= 1, j++)
      if (k & 1)
        x += p[j] << j - 1, y += a << j - 1;
      else
        y += p[j] << j - 1;
    if (x != s1[i] || y != s2[i]) {
      puts("-1");
      return 0;
    }
  }
  for (long long i = 1; i <= a; i++) printf("%lld ", s[i]);
  return 0;
}

```