#include <bits/stdc++.h>
const int maxn = 2e5 + 10;
using namespace std;
long long gcd(long long p, long long q) { return q == 0 ? p : gcd(q, p % q); }
long long qpow(long long p, long long q) {
  long long f = 1;
  while (q) {
    if (q & 1) f = f * p;
    p = p * p;
    q >>= 1;
  }
  return f;
}
inline long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, t;
long long a[maxn], b[maxn], c[maxn], d[maxn], k[maxn], num[30][maxn], sum;
bool flag;
int main() {
  int i, j;
  flag = true;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) b[i] = read();
  for (i = 1; i <= n; i++) c[i] = read();
  for (i = 1; i <= n; i++) d[i] = b[i] + c[i], sum += d[i];
  sum /= (2 * n);
  for (i = 1; i <= n; i++) a[i] = (d[i] - sum) / n;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= 29; j++)
      if ((a[i] >> j) % 2 == 1) num[j][i] = 1, k[j]++;
  }
  for (i = 1; i <= n; i++) {
    long long tmp1 = 0, tmp2 = 0;
    for (j = 0; j <= 29; j++) {
      if (num[j][i])
        tmp1 += (1 << j) * k[j];
      else
        tmp2 += (1 << j) * k[j];
    }
    if (tmp1 != b[i] && tmp2 != c[i]) {
      flag = false;
      break;
    }
  }
  if (flag) {
    for (i = 1; i <= n; i++) printf("%lld ", a[i]);
  } else
    puts("-1");
  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>
const int maxn = 2e5 + 10;
using namespace std;
long long gcd(long long p, long long q) { return q == 0 ? p : gcd(q, p % q); }
long long qpow(long long p, long long q) {
  long long f = 1;
  while (q) {
    if (q & 1) f = f * p;
    p = p * p;
    q >>= 1;
  }
  return f;
}
inline long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, t;
long long a[maxn], b[maxn], c[maxn], d[maxn], k[maxn], num[30][maxn], sum;
bool flag;
int main() {
  int i, j;
  flag = true;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) b[i] = read();
  for (i = 1; i <= n; i++) c[i] = read();
  for (i = 1; i <= n; i++) d[i] = b[i] + c[i], sum += d[i];
  sum /= (2 * n);
  for (i = 1; i <= n; i++) a[i] = (d[i] - sum) / n;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= 29; j++)
      if ((a[i] >> j) % 2 == 1) num[j][i] = 1, k[j]++;
  }
  for (i = 1; i <= n; i++) {
    long long tmp1 = 0, tmp2 = 0;
    for (j = 0; j <= 29; j++) {
      if (num[j][i])
        tmp1 += (1 << j) * k[j];
      else
        tmp2 += (1 << j) * k[j];
    }
    if (tmp1 != b[i] && tmp2 != c[i]) {
      flag = false;
      break;
    }
  }
  if (flag) {
    for (i = 1; i <= n; i++) printf("%lld ", a[i]);
  } else
    puts("-1");
  return 0;
}

```