#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int IT_MAX = 1 << 19;
const int MOD = 1000000007;
const int INF = 1034567891;
const long long LL_INF = 1234567890123456789ll;
const double PI = 3.141592653589793238;
const double ERR = 1E-10;
long long B[200050];
long long C[200050];
long long A[200050];
long long cnt[32];
int main() {
  int N, i, j;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) scanf("%lld", &B[i]);
  for (i = 1; i <= N; i++) scanf("%lld", &C[i]);
  if (N == 1) {
    if (B[1] != C[1])
      return !printf("-1\n");
    else
      return !printf("%lld\n", B[1]);
  }
  long long s = 0;
  for (i = 1; i <= N; i++) s += B[i] + C[i];
  if (s % (2 * N)) return !printf("-1\n");
  s /= 2 * N;
  for (i = 1; i <= N; i++) {
    long long v = B[i] + C[i] - s;
    if (v % N || v < 0) return !printf("-1\n");
    A[i] = v / N;
    if (A[i] > INF) return !printf("-1\n");
  }
  for (i = 1; i <= N; i++)
    for (j = 0; j <= 30; j++)
      if (A[i] & (1 << j)) cnt[j]++;
  for (i = 1; i <= N; i++) {
    long long s = 0;
    for (j = 0; j <= 30; j++) {
      if (A[i] & (1 << j)) s += cnt[j] * (1 << j);
    }
    if (s != B[i]) return !printf("-1\n");
  }
  for (i = 1; i <= N; i++) {
    long long s = 0;
    for (j = 0; j <= 30; j++) {
      if (A[i] & (1 << j))
        s += N * (1 << j);
      else
        s += cnt[j] * (1 << j);
    }
    if (s != C[i]) return !printf("-1\n");
  }
  for (i = 1; i <= N; i++) printf("%lld ", A[i]);
  return !printf("\n");
}

### Prompt

Your task is to create 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>
#pragma warning(disable : 4996)
using namespace std;
int IT_MAX = 1 << 19;
const int MOD = 1000000007;
const int INF = 1034567891;
const long long LL_INF = 1234567890123456789ll;
const double PI = 3.141592653589793238;
const double ERR = 1E-10;
long long B[200050];
long long C[200050];
long long A[200050];
long long cnt[32];
int main() {
  int N, i, j;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) scanf("%lld", &B[i]);
  for (i = 1; i <= N; i++) scanf("%lld", &C[i]);
  if (N == 1) {
    if (B[1] != C[1])
      return !printf("-1\n");
    else
      return !printf("%lld\n", B[1]);
  }
  long long s = 0;
  for (i = 1; i <= N; i++) s += B[i] + C[i];
  if (s % (2 * N)) return !printf("-1\n");
  s /= 2 * N;
  for (i = 1; i <= N; i++) {
    long long v = B[i] + C[i] - s;
    if (v % N || v < 0) return !printf("-1\n");
    A[i] = v / N;
    if (A[i] > INF) return !printf("-1\n");
  }
  for (i = 1; i <= N; i++)
    for (j = 0; j <= 30; j++)
      if (A[i] & (1 << j)) cnt[j]++;
  for (i = 1; i <= N; i++) {
    long long s = 0;
    for (j = 0; j <= 30; j++) {
      if (A[i] & (1 << j)) s += cnt[j] * (1 << j);
    }
    if (s != B[i]) return !printf("-1\n");
  }
  for (i = 1; i <= N; i++) {
    long long s = 0;
    for (j = 0; j <= 30; j++) {
      if (A[i] & (1 << j))
        s += N * (1 << j);
      else
        s += cnt[j] * (1 << j);
    }
    if (s != C[i]) return !printf("-1\n");
  }
  for (i = 1; i <= N; i++) printf("%lld ", A[i]);
  return !printf("\n");
}

```