#include <bits/stdc++.h>
using namespace std;
const int iinf = 1e9 + 7;
const long long linf = 1ll << 60;
const double dinf = 1e10;
template <typename T>
inline void scf(T &x) {
  bool f = 0;
  x = 0;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') {
    f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (f) x = -x;
  return;
}
template <typename T1, typename T2>
void scf(T1 &x, T2 &y) {
  scf(x);
  return scf(y);
}
template <typename T1, typename T2, typename T3>
void scf(T1 &x, T2 &y, T3 &z) {
  scf(x);
  scf(y);
  return scf(z);
}
const int N = 2e5 + 100;
int n;
long long a[N], b[N], c[N], sum;
long long d[30], bb[N], cc[N];
int main() {
  scf(n);
  for (int i = (1); i <= (n); i++) scf(b[i]), sum += b[i];
  for (int i = (1); i <= (n); i++) scf(c[i]), sum += c[i];
  if ((sum % (2 * n)) != 0) {
    puts("-1");
    return 0;
  }
  sum /= (2 * n);
  for (int i = (1); i <= (n); i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] < 0 || a[i] % n != 0) {
      puts("-1");
      return 0;
    }
    a[i] /= n;
    for (int j = 0; j < (30); j++)
      if (a[i] & (1ll << j)) d[j] += (1ll << j);
  }
  for (int i = (1); i <= (n); i++) {
    for (int j = 0; j < (30); j++)
      if (a[i] & (1ll << j))
        bb[i] += d[j];
      else
        cc[i] += d[j];
    cc[i] += a[i] * (long long)n;
    if (bb[i] != b[i] || cc[i] != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = (1); i <= (n); i++) printf("%I64d%c", a[i], i == n ? '\n' : ' ');
  return 0;
}

### Prompt

Please create a solution in cpp 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 iinf = 1e9 + 7;
const long long linf = 1ll << 60;
const double dinf = 1e10;
template <typename T>
inline void scf(T &x) {
  bool f = 0;
  x = 0;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') {
    f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (f) x = -x;
  return;
}
template <typename T1, typename T2>
void scf(T1 &x, T2 &y) {
  scf(x);
  return scf(y);
}
template <typename T1, typename T2, typename T3>
void scf(T1 &x, T2 &y, T3 &z) {
  scf(x);
  scf(y);
  return scf(z);
}
const int N = 2e5 + 100;
int n;
long long a[N], b[N], c[N], sum;
long long d[30], bb[N], cc[N];
int main() {
  scf(n);
  for (int i = (1); i <= (n); i++) scf(b[i]), sum += b[i];
  for (int i = (1); i <= (n); i++) scf(c[i]), sum += c[i];
  if ((sum % (2 * n)) != 0) {
    puts("-1");
    return 0;
  }
  sum /= (2 * n);
  for (int i = (1); i <= (n); i++) {
    a[i] = b[i] + c[i] - sum;
    if (a[i] < 0 || a[i] % n != 0) {
      puts("-1");
      return 0;
    }
    a[i] /= n;
    for (int j = 0; j < (30); j++)
      if (a[i] & (1ll << j)) d[j] += (1ll << j);
  }
  for (int i = (1); i <= (n); i++) {
    for (int j = 0; j < (30); j++)
      if (a[i] & (1ll << j))
        bb[i] += d[j];
      else
        cc[i] += d[j];
    cc[i] += a[i] * (long long)n;
    if (bb[i] != b[i] || cc[i] != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = (1); i <= (n); i++) printf("%I64d%c", a[i], i == n ? '\n' : ' ');
  return 0;
}

```