#include <bits/stdc++.h>
const int NMAX = 200000;
int B[NMAX], C[NMAX], D[NMAX];
int A[NMAX];
bool bits[NMAX][31];
int tot_bits[31];
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", B + i);
  for (int i = 0; i < n; ++i) scanf("%d", C + i);
  for (int i = 0; i < n; ++i) D[i] = B[i] + C[i];
  long long s = 0;
  for (int i = 0; i < n; ++i) s += D[i];
  if (0 != s % (n + n)) {
    printf("-1\n");
    return 0;
  }
  s = s / (n + n);
  for (int i = 0; i < n; ++i) {
    if (0 != (D[i] - s) % n) {
      printf("-1\n");
      return 0;
    }
    A[i] = (D[i] - s) / n;
  }
  for (int i = 0; i < n; ++i) {
    if (A[i] < 0) {
      printf("-1\n");
      return 0;
    }
    for (int j = 0; j < 31; ++j) {
      bits[i][j] = A[i] & (1 << j);
    }
  }
  for (int i = 0; i < 31; ++i) {
    for (int j = 0; j < n; ++j) tot_bits[i] += (bits[j][i] ? 1 : 0);
  }
  for (int i = 0; i < n; ++i) {
    int b = 0, c = 0;
    for (int j = 0; j < 31; ++j) {
      int tmp_b = bits[i][j] ? tot_bits[j] : 0;
      int tmp_c = bits[i][j] ? n : tot_bits[j];
      b += tmp_b * (1 << j);
      c += tmp_c * (1 << j);
    }
    if ((B[i] != b) || (C[i] != c)) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 0; i < n; ++i) printf("%d ", A[i]);
}

### 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 int NMAX = 200000;
int B[NMAX], C[NMAX], D[NMAX];
int A[NMAX];
bool bits[NMAX][31];
int tot_bits[31];
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", B + i);
  for (int i = 0; i < n; ++i) scanf("%d", C + i);
  for (int i = 0; i < n; ++i) D[i] = B[i] + C[i];
  long long s = 0;
  for (int i = 0; i < n; ++i) s += D[i];
  if (0 != s % (n + n)) {
    printf("-1\n");
    return 0;
  }
  s = s / (n + n);
  for (int i = 0; i < n; ++i) {
    if (0 != (D[i] - s) % n) {
      printf("-1\n");
      return 0;
    }
    A[i] = (D[i] - s) / n;
  }
  for (int i = 0; i < n; ++i) {
    if (A[i] < 0) {
      printf("-1\n");
      return 0;
    }
    for (int j = 0; j < 31; ++j) {
      bits[i][j] = A[i] & (1 << j);
    }
  }
  for (int i = 0; i < 31; ++i) {
    for (int j = 0; j < n; ++j) tot_bits[i] += (bits[j][i] ? 1 : 0);
  }
  for (int i = 0; i < n; ++i) {
    int b = 0, c = 0;
    for (int j = 0; j < 31; ++j) {
      int tmp_b = bits[i][j] ? tot_bits[j] : 0;
      int tmp_c = bits[i][j] ? n : tot_bits[j];
      b += tmp_b * (1 << j);
      c += tmp_c * (1 << j);
    }
    if ((B[i] != b) || (C[i] != c)) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 0; i < n; ++i) printf("%d ", A[i]);
}

```