#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
const int MM = 998244353;
const long double PI = acos(-1);
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  return os << '(' << p.first << "," << p.second << ')';
}
const int N = 200005;
long long n;
long long a[N];
long long b[N];
long long c[N];
long long f[33];
int _runtimeTerror_() {
  cin >> n;
  long long sum = 0;
  for (int i = 1; i <= n; ++i) cin >> b[i], sum += b[i];
  for (int i = 1; i <= n; ++i) cin >> c[i], sum += c[i];
  if (sum % (2 * n)) {
    cout << -1;
    return 0;
  }
  sum /= 2 * n;
  for (int i = 1; i <= n; ++i) {
    long long x = b[i] + c[i] - sum;
    if (x % n) {
      cout << -1;
      return 0;
    }
    a[i] = x / n;
    for (int j = 0; j <= 32; ++j)
      if (a[i] & (1LL << j)) f[j]++;
  }
  for (int i = 1; i <= n; ++i) {
    long long x = 0, y = 0;
    for (int j = 0; j <= 32; ++j)
      if (a[i] & (1LL << j)) x += (1LL << j) * f[j];
    for (int j = 0; j <= 32; ++j) {
      if (a[i] & (1LL << j))
        y += n * (1LL << j);
      else
        y += f[j] * (1LL << j);
    }
    if (x != b[i] || y != c[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n; ++i) cout << a[i] << " ";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  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 M = 1000000007;
const int MM = 998244353;
const long double PI = acos(-1);
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  return os << '(' << p.first << "," << p.second << ')';
}
const int N = 200005;
long long n;
long long a[N];
long long b[N];
long long c[N];
long long f[33];
int _runtimeTerror_() {
  cin >> n;
  long long sum = 0;
  for (int i = 1; i <= n; ++i) cin >> b[i], sum += b[i];
  for (int i = 1; i <= n; ++i) cin >> c[i], sum += c[i];
  if (sum % (2 * n)) {
    cout << -1;
    return 0;
  }
  sum /= 2 * n;
  for (int i = 1; i <= n; ++i) {
    long long x = b[i] + c[i] - sum;
    if (x % n) {
      cout << -1;
      return 0;
    }
    a[i] = x / n;
    for (int j = 0; j <= 32; ++j)
      if (a[i] & (1LL << j)) f[j]++;
  }
  for (int i = 1; i <= n; ++i) {
    long long x = 0, y = 0;
    for (int j = 0; j <= 32; ++j)
      if (a[i] & (1LL << j)) x += (1LL << j) * f[j];
    for (int j = 0; j <= 32; ++j) {
      if (a[i] & (1LL << j))
        y += n * (1LL << j);
      else
        y += f[j] * (1LL << j);
    }
    if (x != b[i] || y != c[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n; ++i) cout << a[i] << " ";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  return 0;
}

```