#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T gi() {
  T f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const int INF = 0x3f3f3f3f, N = 200003, M = N << 1;
int n;
long long a[N], b[N], c[N], s[N], sum;
int cnt[35];
inline void nosol() {
  puts("-1");
  exit(0);
}
int main() {
  n = gi<int>();
  for (int i = 1; i <= n; i += 1) b[i] = gi<long long>();
  for (int i = 1; i <= n; i += 1)
    c[i] = gi<long long>(), s[i] = b[i] + c[i], sum += s[i];
  if (sum % (2 * n)) nosol();
  sum /= (2 * n);
  for (int i = 1; i <= n; i += 1) {
    long long na = s[i] - sum;
    if (na % n) nosol();
    na /= n;
    a[i] = na;
  }
  for (int i = 1; i <= n; i += 1)
    for (int j = 0; j <= 32; j += 1)
      if (a[i] >> j & 1ll) ++cnt[j];
  for (int i = 1; i <= n; i += 1) {
    long long s1 = 0;
    for (int j = 0; j <= 32; j += 1)
      if (a[i] >> j & 1ll) s1 += (1ll << j) * cnt[j];
    if (s1 != b[i]) nosol();
  }
  for (int i = 1; i <= n; i += 1) {
    long long s1 = 0;
    for (int j = 0; j <= 32; j += 1)
      if (a[i] >> j & 1ll)
        s1 += (1ll << j) * n;
      else
        s1 += (1ll << j) * cnt[j];
    if (s1 != c[i]) nosol();
  }
  for (int i = 1; i <= n; i += 1) printf("%lld ", a[i]);
  puts("");
  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;
template <typename T>
inline T gi() {
  T f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const int INF = 0x3f3f3f3f, N = 200003, M = N << 1;
int n;
long long a[N], b[N], c[N], s[N], sum;
int cnt[35];
inline void nosol() {
  puts("-1");
  exit(0);
}
int main() {
  n = gi<int>();
  for (int i = 1; i <= n; i += 1) b[i] = gi<long long>();
  for (int i = 1; i <= n; i += 1)
    c[i] = gi<long long>(), s[i] = b[i] + c[i], sum += s[i];
  if (sum % (2 * n)) nosol();
  sum /= (2 * n);
  for (int i = 1; i <= n; i += 1) {
    long long na = s[i] - sum;
    if (na % n) nosol();
    na /= n;
    a[i] = na;
  }
  for (int i = 1; i <= n; i += 1)
    for (int j = 0; j <= 32; j += 1)
      if (a[i] >> j & 1ll) ++cnt[j];
  for (int i = 1; i <= n; i += 1) {
    long long s1 = 0;
    for (int j = 0; j <= 32; j += 1)
      if (a[i] >> j & 1ll) s1 += (1ll << j) * cnt[j];
    if (s1 != b[i]) nosol();
  }
  for (int i = 1; i <= n; i += 1) {
    long long s1 = 0;
    for (int j = 0; j <= 32; j += 1)
      if (a[i] >> j & 1ll)
        s1 += (1ll << j) * n;
      else
        s1 += (1ll << j) * cnt[j];
    if (s1 != c[i]) nosol();
  }
  for (int i = 1; i <= n; i += 1) printf("%lld ", a[i]);
  puts("");
  return 0;
}

```