#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
inline long long read() {
  long long w = 0, r = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') r = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    w = (w << 1) + (w << 3) + (ch ^ 48);
    ch = getchar();
  }
  return w * r;
}
int n;
int bt[N][32], cnt[32][2];
long long suma, sumb, sumc;
long long a[N], cpa[N], b[N], c[N];
void solve() {
  if ((sumb + sumc) % 2 * n != 0) {
    puts("-1");
    return;
  }
  suma = (sumb + sumc) / (2 * n);
  for (int i = 1; i <= n; i++) {
    if ((b[i] + c[i] - suma) % n != 0) {
      puts("-1");
      return;
    }
    cpa[i] = a[i] = (b[i] + c[i] - suma) / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 31; j++) {
      bt[i][j] = cpa[i] & 1;
      cnt[j][bt[i][j]]++;
      cpa[i] >>= 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long tb = 0, tc = 0;
    for (int j = 0; j <= 31; j++) {
      if (bt[i][j] == 1) {
        tb += cnt[j][1] * (1 << j);
        tc += n * (1 << j);
      } else {
        tc += cnt[j][1] * (1 << j);
      }
    }
    if (tb != b[i] || tc != c[i]) {
      puts("-1");
      return;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
  puts("");
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    b[i] = read();
    sumb += b[i];
  }
  for (int i = 1; i <= n; i++) {
    c[i] = read();
    sumc += c[i];
  }
  solve();
  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>
using namespace std;
const int N = 2e5 + 10;
inline long long read() {
  long long w = 0, r = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') r = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    w = (w << 1) + (w << 3) + (ch ^ 48);
    ch = getchar();
  }
  return w * r;
}
int n;
int bt[N][32], cnt[32][2];
long long suma, sumb, sumc;
long long a[N], cpa[N], b[N], c[N];
void solve() {
  if ((sumb + sumc) % 2 * n != 0) {
    puts("-1");
    return;
  }
  suma = (sumb + sumc) / (2 * n);
  for (int i = 1; i <= n; i++) {
    if ((b[i] + c[i] - suma) % n != 0) {
      puts("-1");
      return;
    }
    cpa[i] = a[i] = (b[i] + c[i] - suma) / n;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 31; j++) {
      bt[i][j] = cpa[i] & 1;
      cnt[j][bt[i][j]]++;
      cpa[i] >>= 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    long long tb = 0, tc = 0;
    for (int j = 0; j <= 31; j++) {
      if (bt[i][j] == 1) {
        tb += cnt[j][1] * (1 << j);
        tc += n * (1 << j);
      } else {
        tc += cnt[j][1] * (1 << j);
      }
    }
    if (tb != b[i] || tc != c[i]) {
      puts("-1");
      return;
    }
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
  puts("");
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    b[i] = read();
    sumb += b[i];
  }
  for (int i = 1; i <= n; i++) {
    c[i] = read();
    sumc += c[i];
  }
  solve();
  return 0;
}

```