#include <bits/stdc++.h>
using namespace std;
const int MAX = 200000, MAX_BIT = 31;
int A[MAX], B[MAX], C[MAX];
long long D[MAX];
int B2[MAX], C2[MAX];
int BK[MAX][MAX_BIT], CK[MAX][MAX_BIT];
int K[MAX_BIT];
void solve() {
  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];
  memset(B2, 0, sizeof(B2));
  memset(C2, 0, sizeof(C2));
  memset(K, 0, sizeof(K));
  long long sumA = 0;
  for (int i = 0; i < n; i++) sumA += D[i];
  sumA /= 2 * n;
  for (int i = 0; i < n; i++) A[i] = (D[i] - sumA) / n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < MAX_BIT; j++) {
      if (A[i] & (1 << j)) K[j] += 1;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < MAX_BIT; j++) {
      int a = A[i] & (1 << j);
      if (a >= 1) a = 1;
      BK[i][j] = a == 0 ? 0 : K[j];
      CK[i][j] = a == 0 ? K[j] : n;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < MAX_BIT; j++) {
      B2[i] += BK[i][j] * (1 << j);
      C2[i] += CK[i][j] * (1 << j);
    }
  }
  for (int i = 0; i < n; i++) {
    if (B[i] != B2[i] || C[i] != C2[i]) {
      puts("-1");
      return;
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", A[i]);
  }
  puts("");
}
int main() {
  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 MAX = 200000, MAX_BIT = 31;
int A[MAX], B[MAX], C[MAX];
long long D[MAX];
int B2[MAX], C2[MAX];
int BK[MAX][MAX_BIT], CK[MAX][MAX_BIT];
int K[MAX_BIT];
void solve() {
  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];
  memset(B2, 0, sizeof(B2));
  memset(C2, 0, sizeof(C2));
  memset(K, 0, sizeof(K));
  long long sumA = 0;
  for (int i = 0; i < n; i++) sumA += D[i];
  sumA /= 2 * n;
  for (int i = 0; i < n; i++) A[i] = (D[i] - sumA) / n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < MAX_BIT; j++) {
      if (A[i] & (1 << j)) K[j] += 1;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < MAX_BIT; j++) {
      int a = A[i] & (1 << j);
      if (a >= 1) a = 1;
      BK[i][j] = a == 0 ? 0 : K[j];
      CK[i][j] = a == 0 ? K[j] : n;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < MAX_BIT; j++) {
      B2[i] += BK[i][j] * (1 << j);
      C2[i] += CK[i][j] * (1 << j);
    }
  }
  for (int i = 0; i < n; i++) {
    if (B[i] != B2[i] || C[i] != C2[i]) {
      puts("-1");
      return;
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", A[i]);
  }
  puts("");
}
int main() {
  solve();
  return 0;
}

```