#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5;
int a[MAXN];
int b[MAXN];
int c[MAXN];
int d[MAXN];
long long s_d;
int s_a;
int k[32];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &b[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &c[i]);
  }
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    s_d += d[i];
  }
  s_a = s_d / (2 * n);
  for (int i = 0; i < n; i++) {
    a[i] = d[i] - s_a;
    a[i] /= n;
    if (a[i] < 0) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    int t = a[i];
    for (int j = 0; j < 32; j++) {
      k[j] += (t >> j) % 2;
    }
  }
  for (int i = 0; i < n; i++) {
    int b_t = 0;
    for (int j = 0; j < 32; j++) {
      int bit = (a[i] >> j) & 1;
      if (bit) b_t += (k[j] << j);
    }
    if (b[i] != b_t) {
      printf("-1\n");
      return 0;
    }
    int c_t = 0;
    for (int j = 0; j < 32; j++) {
      int bit = (a[i] >> j) & 1;
      if (bit)
        c_t += (n << j);
      else
        c_t += (k[j] << j);
    }
    if (c[i] != c_t) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%d", a[i]);
    if (i != n - 1) printf(" ");
  }
  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 int MAXN = 2e5;
int a[MAXN];
int b[MAXN];
int c[MAXN];
int d[MAXN];
long long s_d;
int s_a;
int k[32];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &b[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &c[i]);
  }
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    s_d += d[i];
  }
  s_a = s_d / (2 * n);
  for (int i = 0; i < n; i++) {
    a[i] = d[i] - s_a;
    a[i] /= n;
    if (a[i] < 0) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    int t = a[i];
    for (int j = 0; j < 32; j++) {
      k[j] += (t >> j) % 2;
    }
  }
  for (int i = 0; i < n; i++) {
    int b_t = 0;
    for (int j = 0; j < 32; j++) {
      int bit = (a[i] >> j) & 1;
      if (bit) b_t += (k[j] << j);
    }
    if (b[i] != b_t) {
      printf("-1\n");
      return 0;
    }
    int c_t = 0;
    for (int j = 0; j < 32; j++) {
      int bit = (a[i] >> j) & 1;
      if (bit)
        c_t += (n << j);
      else
        c_t += (k[j] << j);
    }
    if (c[i] != c_t) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%d", a[i]);
    if (i != n - 1) printf(" ");
  }
  printf("\n");
  return 0;
}

```