#include <bits/stdc++.h>
using namespace std;
const int maxn = 200500;
const int inf = 2e9;
const double eps = 1e-8;
const int base = 1073676287;
long long a[maxn];
int b[maxn];
int c[maxn];
int bit[64];
int bits2[64];
void check(int n) {
  long long sum = 0LL;
  for (int j = 0; j < n; j++) {
    sum += a[j];
    int x = a[j];
    int pos = 0;
    while (x) {
      bit[pos++] += x % 2;
      x /= 2;
    }
  }
  for (int j = 0; j < n; j++) {
    int x = a[j];
    int pos = 0;
    for (int i = 0; i < 64; i++) bits2[i] = 0;
    while (x) {
      bits2[pos++] = x % 2;
      x /= 2;
    }
    long long cnt = 0LL;
    long long deg = 1LL;
    for (int i = 0; i < 60; i++) {
      if (bits2[i]) cnt += 1LL * bit[i] * deg;
      deg *= 2LL;
    }
    if (cnt != b[j]) {
      puts("-1");
      exit(0);
    }
    if (1LL * a[j] * n + sum != 1LL * b[j] + 1LL * c[j]) {
      puts("-1");
      exit(0);
    }
  }
}
int main() {
  srand(time(0));
  int n;
  long long sum = 0LL;
  scanf("%d", &n);
  for (int j = 0; j < n; j++) {
    scanf("%d", &b[j]);
    sum += 1LL * b[j];
  }
  for (int j = 0; j < n; j++) {
    scanf("%d", &c[j]);
    sum += 1LL * c[j];
  }
  if (sum % (2LL * n) != 0LL) {
    puts("-1");
    return 0;
  }
  sum /= 2LL * n;
  for (int j = 0; j < n; j++) {
    int cur = b[j] + c[j];
    a[j] = cur - sum;
    if (a[j] < 0LL) {
      puts("-1");
      return 0;
    }
    if (1LL * a[j] % n != 0LL) {
      puts("-1");
      return 0;
    }
    a[j] /= 1LL * n;
  }
  check(n);
  for (int j = 0; j < n; j++) cout << a[j] << ' ';
  return 0;
}

### Prompt

Please formulate 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 maxn = 200500;
const int inf = 2e9;
const double eps = 1e-8;
const int base = 1073676287;
long long a[maxn];
int b[maxn];
int c[maxn];
int bit[64];
int bits2[64];
void check(int n) {
  long long sum = 0LL;
  for (int j = 0; j < n; j++) {
    sum += a[j];
    int x = a[j];
    int pos = 0;
    while (x) {
      bit[pos++] += x % 2;
      x /= 2;
    }
  }
  for (int j = 0; j < n; j++) {
    int x = a[j];
    int pos = 0;
    for (int i = 0; i < 64; i++) bits2[i] = 0;
    while (x) {
      bits2[pos++] = x % 2;
      x /= 2;
    }
    long long cnt = 0LL;
    long long deg = 1LL;
    for (int i = 0; i < 60; i++) {
      if (bits2[i]) cnt += 1LL * bit[i] * deg;
      deg *= 2LL;
    }
    if (cnt != b[j]) {
      puts("-1");
      exit(0);
    }
    if (1LL * a[j] * n + sum != 1LL * b[j] + 1LL * c[j]) {
      puts("-1");
      exit(0);
    }
  }
}
int main() {
  srand(time(0));
  int n;
  long long sum = 0LL;
  scanf("%d", &n);
  for (int j = 0; j < n; j++) {
    scanf("%d", &b[j]);
    sum += 1LL * b[j];
  }
  for (int j = 0; j < n; j++) {
    scanf("%d", &c[j]);
    sum += 1LL * c[j];
  }
  if (sum % (2LL * n) != 0LL) {
    puts("-1");
    return 0;
  }
  sum /= 2LL * n;
  for (int j = 0; j < n; j++) {
    int cur = b[j] + c[j];
    a[j] = cur - sum;
    if (a[j] < 0LL) {
      puts("-1");
      return 0;
    }
    if (1LL * a[j] % n != 0LL) {
      puts("-1");
      return 0;
    }
    a[j] /= 1LL * n;
  }
  check(n);
  for (int j = 0; j < n; j++) cout << a[j] << ' ';
  return 0;
}

```