#include <bits/stdc++.h>
#pragma comment(linker, "/stack:640000000")
using namespace std;
int fx[] = {0, 0, -1, 1, -1, 1, 1, -1};
int fy[] = {1, -1, 0, 0, 1, 1, -1, -1};
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
const int N = (int)3e5;
int a[N], b[N], c[N], d[N], n, b1[N], c1[N];
int bits[32][N], k[32];
bool valid() {
  memset(bits, 0, sizeof(bits));
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      if (a[j] < 0) return false;
      if (a[j] & (1 << i))
        bits[i][j] = 1;
      else
        bits[i][j] = 0;
    }
  }
  memset(k, 0, sizeof(k));
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      k[i] += bits[i][j];
    }
  }
  memset(b1, 0, sizeof(b1));
  memset(c1, 0, sizeof(c1));
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      int _b = bits[i][j] ? k[i] : 0;
      int _c = bits[i][j] ? n : k[i];
      b1[j] += _b << i;
      c1[j] += _c << i;
    }
  }
  for (int i = 0; i < n; i++) {
    if (b[i] != b1[i] || c[i] != c1[i]) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%d", &c[i]);
  long long int sum = 0LL;
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    sum += (long long int)d[i];
  }
  sum /= (2 * n);
  for (int i = 0; i < n; i++) a[i] = (d[i] - sum) / n;
  if (!valid())
    printf("-1\n");
  else {
    for (int i = 0; i < n; i++) printf("%d ", a[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 comment(linker, "/stack:640000000")
using namespace std;
int fx[] = {0, 0, -1, 1, -1, 1, 1, -1};
int fy[] = {1, -1, 0, 0, 1, 1, -1, -1};
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
const int N = (int)3e5;
int a[N], b[N], c[N], d[N], n, b1[N], c1[N];
int bits[32][N], k[32];
bool valid() {
  memset(bits, 0, sizeof(bits));
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      if (a[j] < 0) return false;
      if (a[j] & (1 << i))
        bits[i][j] = 1;
      else
        bits[i][j] = 0;
    }
  }
  memset(k, 0, sizeof(k));
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      k[i] += bits[i][j];
    }
  }
  memset(b1, 0, sizeof(b1));
  memset(c1, 0, sizeof(c1));
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      int _b = bits[i][j] ? k[i] : 0;
      int _c = bits[i][j] ? n : k[i];
      b1[j] += _b << i;
      c1[j] += _c << i;
    }
  }
  for (int i = 0; i < n; i++) {
    if (b[i] != b1[i] || c[i] != c1[i]) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%d", &c[i]);
  long long int sum = 0LL;
  for (int i = 0; i < n; i++) {
    d[i] = b[i] + c[i];
    sum += (long long int)d[i];
  }
  sum /= (2 * n);
  for (int i = 0; i < n; i++) a[i] = (d[i] - sum) / n;
  if (!valid())
    printf("-1\n");
  else {
    for (int i = 0; i < n; i++) printf("%d ", a[i]);
    printf("\n");
  }
  return 0;
}

```