#include <bits/stdc++.h>
using namespace std;
const int MX = 2e5 + 5;
const int INF = 0x3f3f3f3f;
int n;
long long tot;
long long B[MX], C[MX], A[MX];
bool solve() {
  if (tot % (2 * n) != 0) return 0;
  tot /= 2 * n;
  for (int i = 1; i <= n; i++) {
    long long now = B[i] + C[i];
    if (now - tot < 0 || (now - tot) % n != 0) return 0;
    A[i] = (now - tot) / n;
  }
  return 1;
}
bool check() {
  int cnt[50] = {0}, P = 33;
  for (int i = 1; i <= n; i++) {
    for (int j = P; j >= 0; j--) {
      if (A[i] >> j & 1) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long nowb = 0, nowc = 0;
    for (int j = P; j >= 0; j--) {
      if (A[i] >> j & 1) {
        nowb += (long long)cnt[j] * (1LL << j);
        nowc += (long long)n * (1LL << j);
      } else
        nowc += (long long)cnt[j] * (1LL << j);
    }
    if (nowb != B[i] || nowc != C[i]) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &B[i]);
    tot += B[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &C[i]);
    tot += C[i];
  }
  if (solve() && check()) {
    for (int i = 1; i <= n; i++) {
      printf("%I64d%c", A[i], i == n ? '\n' : ' ');
    }
  } else
    printf("-1\n");
  return 0;
}

### 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;
const int MX = 2e5 + 5;
const int INF = 0x3f3f3f3f;
int n;
long long tot;
long long B[MX], C[MX], A[MX];
bool solve() {
  if (tot % (2 * n) != 0) return 0;
  tot /= 2 * n;
  for (int i = 1; i <= n; i++) {
    long long now = B[i] + C[i];
    if (now - tot < 0 || (now - tot) % n != 0) return 0;
    A[i] = (now - tot) / n;
  }
  return 1;
}
bool check() {
  int cnt[50] = {0}, P = 33;
  for (int i = 1; i <= n; i++) {
    for (int j = P; j >= 0; j--) {
      if (A[i] >> j & 1) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long nowb = 0, nowc = 0;
    for (int j = P; j >= 0; j--) {
      if (A[i] >> j & 1) {
        nowb += (long long)cnt[j] * (1LL << j);
        nowc += (long long)n * (1LL << j);
      } else
        nowc += (long long)cnt[j] * (1LL << j);
    }
    if (nowb != B[i] || nowc != C[i]) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &B[i]);
    tot += B[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &C[i]);
    tot += C[i];
  }
  if (solve() && check()) {
    for (int i = 1; i <= n; i++) {
      printf("%I64d%c", A[i], i == n ? '\n' : ' ');
    }
  } else
    printf("-1\n");
  return 0;
}

```