#include <bits/stdc++.h>
using namespace std;
inline long long getnum() {
  char c = getchar();
  long long num, sign = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') sign = -1;
  for (num = 0; c >= '0' && c <= '9';) {
    c -= '0';
    num = num * 10 + c;
    c = getchar();
  }
  return num * sign;
}
long long B[200004], C[200004];
long long A[200004], cnt[35];
int main() {
  long long n = getnum();
  long long tot = 0;
  for (int i = 1; i <= n; i++) {
    B[i] = getnum();
    tot += B[i];
  }
  for (int i = 1; i <= n; i++) {
    C[i] = getnum();
    tot += C[i];
  }
  if (tot % (n * 2) != 0) {
    puts("-1");
    return 0;
  }
  tot = tot / (n * 2);
  for (int i = 1; i <= n; i++) {
    A[i] = (B[i] + C[i] - tot) / n;
    if ((B[i] + C[i] - tot) % n != 0) {
      puts("-1");
      return 0;
    }
    if (A[i] < 0) {
      puts("-1");
      return 0;
    }
    long long x = A[i], ind = 1;
    while (x > 0) {
      if (x % 2 == 1) cnt[ind]++;
      x /= 2, ind++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long x = A[i], p = 1, ind = 1;
    while (x > 0) {
      if (x % 2 == 1) B[i] -= p * cnt[ind];
      x /= 2, ind++, p *= 2;
    }
    if (B[i] != 0) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long x = A[i], p = 1, ind = 1;
    while (ind <= 30) {
      if (x % 2 == 1)
        C[i] -= p * n;
      else
        C[i] -= p * cnt[ind];
      x /= 2, ind++, p *= 2;
    }
    if (C[i] != 0) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", A[i]);
}

### 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;
inline long long getnum() {
  char c = getchar();
  long long num, sign = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') sign = -1;
  for (num = 0; c >= '0' && c <= '9';) {
    c -= '0';
    num = num * 10 + c;
    c = getchar();
  }
  return num * sign;
}
long long B[200004], C[200004];
long long A[200004], cnt[35];
int main() {
  long long n = getnum();
  long long tot = 0;
  for (int i = 1; i <= n; i++) {
    B[i] = getnum();
    tot += B[i];
  }
  for (int i = 1; i <= n; i++) {
    C[i] = getnum();
    tot += C[i];
  }
  if (tot % (n * 2) != 0) {
    puts("-1");
    return 0;
  }
  tot = tot / (n * 2);
  for (int i = 1; i <= n; i++) {
    A[i] = (B[i] + C[i] - tot) / n;
    if ((B[i] + C[i] - tot) % n != 0) {
      puts("-1");
      return 0;
    }
    if (A[i] < 0) {
      puts("-1");
      return 0;
    }
    long long x = A[i], ind = 1;
    while (x > 0) {
      if (x % 2 == 1) cnt[ind]++;
      x /= 2, ind++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long x = A[i], p = 1, ind = 1;
    while (x > 0) {
      if (x % 2 == 1) B[i] -= p * cnt[ind];
      x /= 2, ind++, p *= 2;
    }
    if (B[i] != 0) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long x = A[i], p = 1, ind = 1;
    while (ind <= 30) {
      if (x % 2 == 1)
        C[i] -= p * n;
      else
        C[i] -= p * cnt[ind];
      x /= 2, ind++, p *= 2;
    }
    if (C[i] != 0) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", A[i]);
}

```