#include <bits/stdc++.h>
const long long N = 2e5 + 7;
long long a[N], b[N], c[N];
long long n, ans[N], f[63];
inline long long read() {
  long long x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x;
}
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
signed main() {
  n = read();
  for (long long i = 1; i <= n; ++i) b[i] = read();
  for (long long i = 1; i <= n; ++i) c[i] = read();
  long long S = 0;
  for (long long i = 1; i <= n; ++i) S += b[i] + c[i];
  if (gcd(S, 2 * n) != 2 * n) {
    puts("-1");
    return 0;
  }
  S /= 2 * n;
  for (long long i = 1; i <= n; ++i) {
    long long x = b[i] + c[i];
    x -= S;
    if (x < 0) {
      puts("-1");
      return 0;
    }
    if (gcd(x, n) != n) {
      puts("-1");
      return 0;
    }
    a[i] = x / n;
  }
  for (long long i = 1; i <= n; ++i) {
    for (long long j = 0; j <= 32; ++j) {
      if (a[i] & (1 << j)) ++f[j];
    }
  }
  for (long long i = 1; i <= n; ++i) {
    long long p = 0, q = 0;
    for (long long j = 0; j <= 31; ++j) {
      if (a[i] & (1 << j))
        p += f[j] * (1 << j), q += n * (1 << j);
      else
        q += f[j] * (1 << j);
    }
    if (p != b[i] || q != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (long long i = 1; i <= n; ++i) {
    printf("%lld ", a[i]);
  }
  puts("");
  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>
const long long N = 2e5 + 7;
long long a[N], b[N], c[N];
long long n, ans[N], f[63];
inline long long read() {
  long long x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x;
}
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
signed main() {
  n = read();
  for (long long i = 1; i <= n; ++i) b[i] = read();
  for (long long i = 1; i <= n; ++i) c[i] = read();
  long long S = 0;
  for (long long i = 1; i <= n; ++i) S += b[i] + c[i];
  if (gcd(S, 2 * n) != 2 * n) {
    puts("-1");
    return 0;
  }
  S /= 2 * n;
  for (long long i = 1; i <= n; ++i) {
    long long x = b[i] + c[i];
    x -= S;
    if (x < 0) {
      puts("-1");
      return 0;
    }
    if (gcd(x, n) != n) {
      puts("-1");
      return 0;
    }
    a[i] = x / n;
  }
  for (long long i = 1; i <= n; ++i) {
    for (long long j = 0; j <= 32; ++j) {
      if (a[i] & (1 << j)) ++f[j];
    }
  }
  for (long long i = 1; i <= n; ++i) {
    long long p = 0, q = 0;
    for (long long j = 0; j <= 31; ++j) {
      if (a[i] & (1 << j))
        p += f[j] * (1 << j), q += n * (1 << j);
      else
        q += f[j] * (1 << j);
    }
    if (p != b[i] || q != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (long long i = 1; i <= n; ++i) {
    printf("%lld ", a[i]);
  }
  puts("");
  return 0;
}

```