#include <bits/stdc++.h>
using namespace std;
int n;
const int MAXN = 200010;
int B[MAXN], C[MAXN], a[MAXN];
long long b[MAXN], c[MAXN];
struct Node {
  int b, c, id;
  bool operator<(const Node ano) const { return b + c < ano.b + ano.c; }
} opt[MAXN];
int cnt[MAXN];
bool check(int x) {
  int now = x;
  for (int i = 1; i <= n; i++) {
    a[opt[i].id] = now;
    now += ((opt[i + 1].b + opt[i + 1].c - opt[i].b - opt[i].c) / n);
  }
  memset(cnt, 0, sizeof cnt);
  memset(b, 0, sizeof b);
  memset(c, 0, sizeof c);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) {
        b[i] += 1LL * cnt[j] * (1 << j);
        c[i] += 1LL * n * (1 << j);
      } else {
        c[i] += 1LL * cnt[j] * (1 << j);
      }
    }
  }
  return b[1] + c[1] >= B[1] + C[1];
}
bool check2(int x) {
  int now = x;
  for (int i = 1; i <= n; i++) {
    a[opt[i].id] = now;
    now += ((opt[i + 1].b + opt[i + 1].c - opt[i].b - opt[i].c) / n);
  }
  memset(cnt, 0, sizeof cnt);
  memset(b, 0, sizeof b);
  memset(c, 0, sizeof c);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) {
        b[i] += 1LL * cnt[j] * (1 << j);
        c[i] += 1LL * n * (1 << j);
      } else {
        c[i] += 1LL * cnt[j] * (1 << j);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if ((b[i] != B[i]) || (c[i] != C[i])) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d", &opt[i].b), B[i] = opt[i].b, opt[i].id = i;
  for (int i = 1; i <= n; i++) scanf("%d", &opt[i].c), C[i] = opt[i].c;
  sort(opt + 1, opt + n + 1);
  for (int i = 2; i <= n; i++) {
    if ((opt[i].b + opt[i].c - opt[i - 1].b - opt[i - 1].c) % n != 0) {
      puts("-1");
      return 0;
    }
  }
  int l = 0, r = 1000000000;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  if (check2(l)) {
    for (int i = 1; i <= n; i++) printf("%d%c", a[i], (i == n ? 10 : ' '));
  } else
    puts("-1");
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
int n;
const int MAXN = 200010;
int B[MAXN], C[MAXN], a[MAXN];
long long b[MAXN], c[MAXN];
struct Node {
  int b, c, id;
  bool operator<(const Node ano) const { return b + c < ano.b + ano.c; }
} opt[MAXN];
int cnt[MAXN];
bool check(int x) {
  int now = x;
  for (int i = 1; i <= n; i++) {
    a[opt[i].id] = now;
    now += ((opt[i + 1].b + opt[i + 1].c - opt[i].b - opt[i].c) / n);
  }
  memset(cnt, 0, sizeof cnt);
  memset(b, 0, sizeof b);
  memset(c, 0, sizeof c);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) {
        b[i] += 1LL * cnt[j] * (1 << j);
        c[i] += 1LL * n * (1 << j);
      } else {
        c[i] += 1LL * cnt[j] * (1 << j);
      }
    }
  }
  return b[1] + c[1] >= B[1] + C[1];
}
bool check2(int x) {
  int now = x;
  for (int i = 1; i <= n; i++) {
    a[opt[i].id] = now;
    now += ((opt[i + 1].b + opt[i + 1].c - opt[i].b - opt[i].c) / n);
  }
  memset(cnt, 0, sizeof cnt);
  memset(b, 0, sizeof b);
  memset(c, 0, sizeof c);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) cnt[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      if ((a[i] >> j) & 1) {
        b[i] += 1LL * cnt[j] * (1 << j);
        c[i] += 1LL * n * (1 << j);
      } else {
        c[i] += 1LL * cnt[j] * (1 << j);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if ((b[i] != B[i]) || (c[i] != C[i])) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d", &opt[i].b), B[i] = opt[i].b, opt[i].id = i;
  for (int i = 1; i <= n; i++) scanf("%d", &opt[i].c), C[i] = opt[i].c;
  sort(opt + 1, opt + n + 1);
  for (int i = 2; i <= n; i++) {
    if ((opt[i].b + opt[i].c - opt[i - 1].b - opt[i - 1].c) % n != 0) {
      puts("-1");
      return 0;
    }
  }
  int l = 0, r = 1000000000;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  if (check2(l)) {
    for (int i = 1; i <= n; i++) printf("%d%c", a[i], (i == n ? 10 : ' '));
  } else
    puts("-1");
}

```