#include <bits/stdc++.h>
using namespace std;
const int N = 200009;
int n, a[N], b[N], c[N];
inline int read() {
  char c = getchar();
  int ret = 0, f = 1;
  while (c < '0' || c > '9') {
    f = c == '-' ? -1 : 1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    ret = ret * 10 + c - '0';
    c = getchar();
  }
  return ret * f;
}
inline bool check() {
  static int cnt[100];
  for (int i = 1; i <= n; i++) {
    for (int t = 0, v = a[i]; v; v >>= 1, ++t) {
      cnt[t] += v & 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    int tb = 0, tc = 0;
    for (int j = 0, cur = 1; j <= 30; j++, cur <<= 1) {
      tb += (a[i] & cur) ? cnt[j] * cur : 0;
      tc += (a[i] & cur) ? n * cur : cnt[j] * cur;
    }
    if (tb != b[i] || tc != c[i]) {
      return false;
    }
  }
  return true;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = b[i] = read();
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    a[i] += (c[i] = read());
    sum += a[i];
  }
  if (sum % (n << 1)) {
    puts("-1");
    exit(0);
  }
  sum /= n << 1;
  for (int i = 1; i <= n; i++) {
    if ((a[i] -= sum) % n) {
      puts("-1");
      exit(0);
    } else {
      a[i] /= n;
    }
  }
  if (!check()) {
    puts("-1");
    exit(0);
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", a[i]);
  }
  return 0;
}

### Prompt

Your challenge is to write 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 N = 200009;
int n, a[N], b[N], c[N];
inline int read() {
  char c = getchar();
  int ret = 0, f = 1;
  while (c < '0' || c > '9') {
    f = c == '-' ? -1 : 1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    ret = ret * 10 + c - '0';
    c = getchar();
  }
  return ret * f;
}
inline bool check() {
  static int cnt[100];
  for (int i = 1; i <= n; i++) {
    for (int t = 0, v = a[i]; v; v >>= 1, ++t) {
      cnt[t] += v & 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    int tb = 0, tc = 0;
    for (int j = 0, cur = 1; j <= 30; j++, cur <<= 1) {
      tb += (a[i] & cur) ? cnt[j] * cur : 0;
      tc += (a[i] & cur) ? n * cur : cnt[j] * cur;
    }
    if (tb != b[i] || tc != c[i]) {
      return false;
    }
  }
  return true;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = b[i] = read();
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    a[i] += (c[i] = read());
    sum += a[i];
  }
  if (sum % (n << 1)) {
    puts("-1");
    exit(0);
  }
  sum /= n << 1;
  for (int i = 1; i <= n; i++) {
    if ((a[i] -= sum) % n) {
      puts("-1");
      exit(0);
    } else {
      a[i] /= n;
    }
  }
  if (!check()) {
    puts("-1");
    exit(0);
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", a[i]);
  }
  return 0;
}

```