#include <bits/stdc++.h>
FILE *in = stdin, *out = stdout;
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:36777216")
using namespace std;
int n;
long long int b[500005], c[500005], k[500005], a[500005];
long long int OR[500005], AND[500005];
int main() {
  fscanf(in, "%d", &n);
  for (int i = (1); i <= (n); i++) fscanf(in, "%lld", &b[i]);
  for (int i = (1); i <= (n); i++) fscanf(in, "%lld", &c[i]);
  long long int sum = 0;
  for (int i = (1); i <= (n); i++) k[i] = b[i] + c[i], sum += k[i];
  if (n == 1) {
    if (c[1] == b[1])
      fprintf(out, "%lld", b[1]);
    else
      fprintf(out, "-1");
    return 0;
  }
  long long int m = k[1] % n;
  for (int i = (2); i <= (n); i++) {
    if (k[i] % n != k[1] % n) {
      fprintf(out, "-1");
      return 0;
    }
  }
  if (sum % (n * 2) != 0) {
    fprintf(out, "-1");
    return 0;
  }
  long long int S = sum / n / 2;
  for (int i = (1); i <= (n); i++) {
    if (k[i] - S < 0) {
      fprintf(out, "-1");
      return 0;
    }
    a[i] = (k[i] - S) / n;
  }
  for (int i = (1); i <= (n); i++) {
    for (int j = (0); j <= (30); j++) {
      if (a[i] & (1 << j)) OR[j]++;
    }
  }
  for (int i = (1); i <= (n); i++) {
    long long int B = 0, C = 0;
    for (int j = (0); j <= (30); j++) {
      if (a[i] & (1 << j))
        B += (1 << j) * OR[j], C += (1 << j) * (long long int)n;
      else
        C += (1 << j) * OR[j];
    }
    if (b[i] != B || c[i] != C) {
      fprintf(out, "-1");
      return 0;
    }
  }
  for (int i = (1); i <= (n); i++) {
    fprintf(out, "%lld ", a[i]);
  }
  return 0;
}

### Prompt

Your task is to create 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>
FILE *in = stdin, *out = stdout;
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:36777216")
using namespace std;
int n;
long long int b[500005], c[500005], k[500005], a[500005];
long long int OR[500005], AND[500005];
int main() {
  fscanf(in, "%d", &n);
  for (int i = (1); i <= (n); i++) fscanf(in, "%lld", &b[i]);
  for (int i = (1); i <= (n); i++) fscanf(in, "%lld", &c[i]);
  long long int sum = 0;
  for (int i = (1); i <= (n); i++) k[i] = b[i] + c[i], sum += k[i];
  if (n == 1) {
    if (c[1] == b[1])
      fprintf(out, "%lld", b[1]);
    else
      fprintf(out, "-1");
    return 0;
  }
  long long int m = k[1] % n;
  for (int i = (2); i <= (n); i++) {
    if (k[i] % n != k[1] % n) {
      fprintf(out, "-1");
      return 0;
    }
  }
  if (sum % (n * 2) != 0) {
    fprintf(out, "-1");
    return 0;
  }
  long long int S = sum / n / 2;
  for (int i = (1); i <= (n); i++) {
    if (k[i] - S < 0) {
      fprintf(out, "-1");
      return 0;
    }
    a[i] = (k[i] - S) / n;
  }
  for (int i = (1); i <= (n); i++) {
    for (int j = (0); j <= (30); j++) {
      if (a[i] & (1 << j)) OR[j]++;
    }
  }
  for (int i = (1); i <= (n); i++) {
    long long int B = 0, C = 0;
    for (int j = (0); j <= (30); j++) {
      if (a[i] & (1 << j))
        B += (1 << j) * OR[j], C += (1 << j) * (long long int)n;
      else
        C += (1 << j) * OR[j];
    }
    if (b[i] != B || c[i] != C) {
      fprintf(out, "-1");
      return 0;
    }
  }
  for (int i = (1); i <= (n); i++) {
    fprintf(out, "%lld ", a[i]);
  }
  return 0;
}

```