#include <bits/stdc++.h>
const double pi = acos(-1.0);
const double eps = 1e-9;
const int inf = 2 * 1e9;
const int base = 2011;
const int raz = (int)2e5 + 5;
const double deg = pi / 180;
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  long long b[n], c[n], sb, sc;
  sb = sc = 0;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &b[i]);
    sb += b[i];
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", &c[i]);
    sc += c[i];
  }
  long long sm = (sb + sc) / (2 * n);
  long long a[n];
  for (int i = 0; i < n; i++) {
    a[i] = (b[i] + c[i] - sm) / n;
    if (a[i] < 0) {
      printf("-1");
      return 0;
    }
  }
  bool bt[31][n];
  long long sbt[31];
  memset(bt, 0, sizeof(bt));
  memset(sbt, 0, sizeof(sbt));
  for (int i = 0; i < 31; i++)
    for (int j = 0; j < n; j++)
      if ((a[j] & (1LL << i))) bt[i][j] = 1;
  for (int i = 0; i < 31; i++) {
    sbt[i] = 0;
    for (int j = 0; j < n; j++) sbt[i] += bt[i][j];
  }
  long long cb, cc;
  for (int i = 0; i < n; i++) {
    cb = cc = 0;
    for (int bit = 0; bit < 31; bit++) {
      if (bt[bit][i]) {
        cb += sbt[bit] << bit;
        cc += n << bit;
      } else
        cc += sbt[bit] << bit;
    }
    if (cb != b[i] or cc != c[i]) {
      printf("-1");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) printf("%lld ", a[i]);
  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>
const double pi = acos(-1.0);
const double eps = 1e-9;
const int inf = 2 * 1e9;
const int base = 2011;
const int raz = (int)2e5 + 5;
const double deg = pi / 180;
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  long long b[n], c[n], sb, sc;
  sb = sc = 0;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &b[i]);
    sb += b[i];
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", &c[i]);
    sc += c[i];
  }
  long long sm = (sb + sc) / (2 * n);
  long long a[n];
  for (int i = 0; i < n; i++) {
    a[i] = (b[i] + c[i] - sm) / n;
    if (a[i] < 0) {
      printf("-1");
      return 0;
    }
  }
  bool bt[31][n];
  long long sbt[31];
  memset(bt, 0, sizeof(bt));
  memset(sbt, 0, sizeof(sbt));
  for (int i = 0; i < 31; i++)
    for (int j = 0; j < n; j++)
      if ((a[j] & (1LL << i))) bt[i][j] = 1;
  for (int i = 0; i < 31; i++) {
    sbt[i] = 0;
    for (int j = 0; j < n; j++) sbt[i] += bt[i][j];
  }
  long long cb, cc;
  for (int i = 0; i < n; i++) {
    cb = cc = 0;
    for (int bit = 0; bit < 31; bit++) {
      if (bt[bit][i]) {
        cb += sbt[bit] << bit;
        cc += n << bit;
      } else
        cc += sbt[bit] << bit;
    }
    if (cb != b[i] or cc != c[i]) {
      printf("-1");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) printf("%lld ", a[i]);
  return 0;
}

```