#include <bits/stdc++.h>
using namespace std;
const int mx = 2e5 + 10;
const int mxb = 60;
long long int a[mx], b[mx], c[mx];
int cnt[mxb][2];
int bat(int idx, long long int v) {
  if (v & (1ll << idx)) return 1;
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  long long int sm = 0;
  for (int i = 0; i < n; i++) scanf("%lld", b + i);
  for (int i = 0; i < n; i++) scanf("%lld", c + i), sm += b[i] + c[i];
  sm /= n + n;
  for (int i = 0; i < n; i++) {
    a[i] = (b[i] + c[i] - sm) / n;
    if (a[i] < 0) {
      puts("-1");
      return 0;
    }
    for (int j = 0; j < mxb; j++) cnt[j][bat(j, a[i])]++;
  }
  for (int i = 0; i < n; i++) {
    long long int bi = 0, ci = 0;
    int cb = 0, cc = 0;
    for (int j = 0; j < mxb; j++) {
      int db = cb, dc = cc;
      for (int x = 0; x < 2; x++) {
        db += cnt[j][x] * (bat(j, a[i]) & x);
        dc += cnt[j][x] * (bat(j, a[i]) | x);
      }
      cb = db / 2, cc = dc / 2;
      if (db & 1) bi |= (1ll << j);
      if (dc & 1) ci |= (1ll << j);
    }
    if (bi != b[i] || ci != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) printf("%lld ", a[i]);
  puts("");
  return 0;
}

### Prompt

In CPP, your task is to solve 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 mx = 2e5 + 10;
const int mxb = 60;
long long int a[mx], b[mx], c[mx];
int cnt[mxb][2];
int bat(int idx, long long int v) {
  if (v & (1ll << idx)) return 1;
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  long long int sm = 0;
  for (int i = 0; i < n; i++) scanf("%lld", b + i);
  for (int i = 0; i < n; i++) scanf("%lld", c + i), sm += b[i] + c[i];
  sm /= n + n;
  for (int i = 0; i < n; i++) {
    a[i] = (b[i] + c[i] - sm) / n;
    if (a[i] < 0) {
      puts("-1");
      return 0;
    }
    for (int j = 0; j < mxb; j++) cnt[j][bat(j, a[i])]++;
  }
  for (int i = 0; i < n; i++) {
    long long int bi = 0, ci = 0;
    int cb = 0, cc = 0;
    for (int j = 0; j < mxb; j++) {
      int db = cb, dc = cc;
      for (int x = 0; x < 2; x++) {
        db += cnt[j][x] * (bat(j, a[i]) & x);
        dc += cnt[j][x] * (bat(j, a[i]) | x);
      }
      cb = db / 2, cc = dc / 2;
      if (db & 1) bi |= (1ll << j);
      if (dc & 1) ci |= (1ll << j);
    }
    if (bi != b[i] || ci != c[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) printf("%lld ", a[i]);
  puts("");
  return 0;
}

```