#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const double EPS = 1e-15;
const double PI = acos(-1.0);
const int MAXN = (int)2e5 + 7;
const int LOGN = 32;
const int INF = (int)1e9 + 7;
const long long LINF = (long long)4e18 + 7;
const int MOD = (int)1e9 + 7;
const unsigned long long P = 239017;
const unsigned long long MM = (unsigned long long)2147482661;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
int solve();
int main() { return solve(); }
long long a[MAXN], b[MAXN], c[MAXN], sumbc;
long long cnt[LOGN + 1], pw[LOGN + 1];
void no() {
  puts("-1");
  exit(0);
}
int get(long long msk, int i) { return (msk >> i) & 1; }
int solve() {
  int n;
  cin >> n;
  for (int i = 0; i < (int)n; i++) scanf("%I64d", b + i);
  for (int i = 0; i < (int)n; i++) scanf("%I64d", c + i);
  for (int i = 0; i < (int)n; i++) sumbc += b[i] + c[i];
  if (sumbc % (2 * n) != 0) no();
  long long suma = sumbc / (2 * n);
  for (int i = 0; i < (int)n; i++) {
    a[i] = (b[i] + c[i] - suma) / n;
    if (a[i] < 0) no();
    if (a[i] * n != (b[i] + c[i] - suma)) no();
  }
  pw[0] = 1ll;
  for (int bit = 0; bit < (int)LOGN; bit++) {
    pw[bit + 1] = (pw[bit] << 1);
    for (int i = 0; i < (int)n; i++) {
      cnt[bit] += get(a[i], bit);
    }
  }
  for (int i = 0; i < (int)n; i++) {
    long long curb = 0, curc = 0;
    for (int bit = 0; bit < (int)LOGN; bit++) {
      if (get(a[i], bit)) {
        curb += cnt[bit] * pw[bit];
        curc += n * pw[bit];
      } else {
        curc += cnt[bit] * pw[bit];
      }
    }
    if (b[i] != curb || c[i] != curc) no();
  }
  for (int i = 0; i < (int)n; i++) {
    printf("%I64d ", a[i]);
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const double EPS = 1e-15;
const double PI = acos(-1.0);
const int MAXN = (int)2e5 + 7;
const int LOGN = 32;
const int INF = (int)1e9 + 7;
const long long LINF = (long long)4e18 + 7;
const int MOD = (int)1e9 + 7;
const unsigned long long P = 239017;
const unsigned long long MM = (unsigned long long)2147482661;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
int solve();
int main() { return solve(); }
long long a[MAXN], b[MAXN], c[MAXN], sumbc;
long long cnt[LOGN + 1], pw[LOGN + 1];
void no() {
  puts("-1");
  exit(0);
}
int get(long long msk, int i) { return (msk >> i) & 1; }
int solve() {
  int n;
  cin >> n;
  for (int i = 0; i < (int)n; i++) scanf("%I64d", b + i);
  for (int i = 0; i < (int)n; i++) scanf("%I64d", c + i);
  for (int i = 0; i < (int)n; i++) sumbc += b[i] + c[i];
  if (sumbc % (2 * n) != 0) no();
  long long suma = sumbc / (2 * n);
  for (int i = 0; i < (int)n; i++) {
    a[i] = (b[i] + c[i] - suma) / n;
    if (a[i] < 0) no();
    if (a[i] * n != (b[i] + c[i] - suma)) no();
  }
  pw[0] = 1ll;
  for (int bit = 0; bit < (int)LOGN; bit++) {
    pw[bit + 1] = (pw[bit] << 1);
    for (int i = 0; i < (int)n; i++) {
      cnt[bit] += get(a[i], bit);
    }
  }
  for (int i = 0; i < (int)n; i++) {
    long long curb = 0, curc = 0;
    for (int bit = 0; bit < (int)LOGN; bit++) {
      if (get(a[i], bit)) {
        curb += cnt[bit] * pw[bit];
        curc += n * pw[bit];
      } else {
        curc += cnt[bit] * pw[bit];
      }
    }
    if (b[i] != curb || c[i] != curc) no();
  }
  for (int i = 0; i < (int)n; i++) {
    printf("%I64d ", a[i]);
  }
  return 0;
}

```