#include <bits/stdc++.h>
using namespace std;
struct FastReader {
  FastReader &operator,(int &x) {
    scanf("%d", &x);
    return *this;
  }
  FastReader &operator,(signed long long &x) {
    scanf(
        "%"
        "ll"
        "d",
        &x);
    return *this;
  }
  FastReader &operator,(double &x) {
    scanf("%lf", &x);
    return *this;
  }
  FastReader &operator,(long double &x) {
    scanf("%Lf", &x);
    return *this;
  }
  FastReader &operator,(char &x) {
    do x = getchar();
    while (x == ' ' || x == '\n');
    return *this;
  }
  FastReader &operator,(char *x) {
    scanf("%s", x);
    return *this;
  }
  FastReader &operator,(string &x) {
    cin >> x;
    return *this;
  }
} in;
struct FastWriter {
  FastWriter &operator,(const int x) {
    printf("%d", x);
    return *this;
  }
  FastWriter &operator,(const signed long long x) {
    printf(
        "%"
        "ll"
        "d",
        x);
    return *this;
  }
  FastWriter &operator,(const double x) {
    printf(
        "%."
        "8"
        "lf",
        x);
    return *this;
  }
  FastWriter &operator,(const long double x) {
    printf(
        "%."
        "12"
        "Lf",
        x);
    return *this;
  }
  FastWriter &operator,(const char x) {
    putchar(x);
    return *this;
  }
  FastWriter &operator,(const char *x) {
    printf("%s", x);
    return *this;
  }
  FastWriter &operator,(const string &x) { return operator,(x.c_str()); }
} out;
template <class T>
inline T sqr(T x) {
  return x * x;
}
static const int N = 2e5 + 111, K = 33;
int n;
signed long long a[N], b[N], c[N], k[K];
signed long long sum;
int main() {
  in, n;
  for (int i = 0; i < n; i++) in, b[i];
  for (int i = 0; i < n; i++) in, c[i];
  for (int i = 0; i < n; i++) sum += b[i] + c[i];
  sum /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = (b[i] + c[i] - sum) / n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < K; j++)
      if (((1LL << j) & a[i]) > 0) k[j]++;
  }
  for (int i = 0; i < n; i++) {
    signed long long sb = 0, sc = 0;
    for (int j = 0; j < K; j++) {
      if (((1LL << j) & a[i]) > 0)
        sb += k[j] * (1LL << j), sc += n * (1LL << j);
      else
        sc += k[j] * (1LL << j);
    }
    if (sb != b[i] || sc != c[i]) {
      out, -1;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) out, a[i], ' ';
  return 0;
}

### Prompt

Create a solution in CPP for 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;
struct FastReader {
  FastReader &operator,(int &x) {
    scanf("%d", &x);
    return *this;
  }
  FastReader &operator,(signed long long &x) {
    scanf(
        "%"
        "ll"
        "d",
        &x);
    return *this;
  }
  FastReader &operator,(double &x) {
    scanf("%lf", &x);
    return *this;
  }
  FastReader &operator,(long double &x) {
    scanf("%Lf", &x);
    return *this;
  }
  FastReader &operator,(char &x) {
    do x = getchar();
    while (x == ' ' || x == '\n');
    return *this;
  }
  FastReader &operator,(char *x) {
    scanf("%s", x);
    return *this;
  }
  FastReader &operator,(string &x) {
    cin >> x;
    return *this;
  }
} in;
struct FastWriter {
  FastWriter &operator,(const int x) {
    printf("%d", x);
    return *this;
  }
  FastWriter &operator,(const signed long long x) {
    printf(
        "%"
        "ll"
        "d",
        x);
    return *this;
  }
  FastWriter &operator,(const double x) {
    printf(
        "%."
        "8"
        "lf",
        x);
    return *this;
  }
  FastWriter &operator,(const long double x) {
    printf(
        "%."
        "12"
        "Lf",
        x);
    return *this;
  }
  FastWriter &operator,(const char x) {
    putchar(x);
    return *this;
  }
  FastWriter &operator,(const char *x) {
    printf("%s", x);
    return *this;
  }
  FastWriter &operator,(const string &x) { return operator,(x.c_str()); }
} out;
template <class T>
inline T sqr(T x) {
  return x * x;
}
static const int N = 2e5 + 111, K = 33;
int n;
signed long long a[N], b[N], c[N], k[K];
signed long long sum;
int main() {
  in, n;
  for (int i = 0; i < n; i++) in, b[i];
  for (int i = 0; i < n; i++) in, c[i];
  for (int i = 0; i < n; i++) sum += b[i] + c[i];
  sum /= 2 * n;
  for (int i = 0; i < n; i++) a[i] = (b[i] + c[i] - sum) / n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < K; j++)
      if (((1LL << j) & a[i]) > 0) k[j]++;
  }
  for (int i = 0; i < n; i++) {
    signed long long sb = 0, sc = 0;
    for (int j = 0; j < K; j++) {
      if (((1LL << j) & a[i]) > 0)
        sb += k[j] * (1LL << j), sc += n * (1LL << j);
      else
        sc += k[j] * (1LL << j);
    }
    if (sb != b[i] || sc != c[i]) {
      out, -1;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) out, a[i], ' ';
  return 0;
}

```