#include <bits/stdc++.h>
using namespace std;
const long long linf = 1e18 + 5;
const int mod = (int)1e9 + 7;
const int inf = 1e9;
long long read() {
  bool minus = false;
  long long result = 0;
  char ch;
  ch = getchar();
  while (true) {
    if (ch == '-') break;
    if (ch >= '0' && ch <= '9') break;
    ch = getchar();
  }
  if (ch == '-')
    minus = true;
  else
    result = ch - '0';
  while (true) {
    ch = getchar();
    if (ch < '0' || ch > '9') break;
    result = result * 10 + (ch - '0');
  }
  if (minus) {
    return -result;
  } else {
    return result;
  }
}
long long fpow(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power % 2 == 1) result = (result * base);
    base = (base * base);
    power /= 2;
  }
  return result;
}
long long n, b[300000], c[300000], a[300000], d[300000], b1[300000], c1[300000];
long long bits[31][300000], kbit[31];
int main() {
  n = read();
  long long i, j, sum = 0;
  for (i = 0; i < n; i++) {
    b[i] = read();
  }
  for (i = 0; i < n; i++) {
    c[i] = read();
    d[i] = b[i] + c[i];
  }
  for (i = 0; i < n; i++) {
    sum += d[i];
  }
  sum /= 2 * n;
  for (i = 0; i < n; i++) {
    a[i] = (d[i] - sum) / n;
    if (a[i] < 0) {
      return 0 * printf("-1\n");
    }
  }
  for (i = 0; i < 31; i++) {
    for (j = 0; j < n; j++) {
      if ((a[j] & (1LL << i)) == 0)
        bits[i][j] = 0;
      else
        bits[i][j] = 1;
    }
  }
  for (i = 0; i < 31; i++) {
    kbit[i] = 0;
    for (j = 0; j < n; j++) {
      kbit[i] += bits[i][j];
    }
  }
  for (i = 0; i < n; i++) {
    b1[i] = 0;
    c1[i] = 0;
  }
  for (i = 0; i < 31; i++) {
    for (j = 0; j < n; j++) {
      int bbs = bits[i][j] ? kbit[i] : 0;
      int cbs = bits[i][j] ? n : kbit[i];
      b1[j] += bbs << i;
      c1[j] += cbs << i;
    }
  }
  for (i = 0; i < n; i++) {
    if (b1[i] != b[i] || c1[i] != c[i]) {
      return 0 * printf("-1\n");
    }
  }
  for (i = 0; i < n; i++) {
    printf("%lld ", a[i]);
  }
  printf("\n");
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
const long long linf = 1e18 + 5;
const int mod = (int)1e9 + 7;
const int inf = 1e9;
long long read() {
  bool minus = false;
  long long result = 0;
  char ch;
  ch = getchar();
  while (true) {
    if (ch == '-') break;
    if (ch >= '0' && ch <= '9') break;
    ch = getchar();
  }
  if (ch == '-')
    minus = true;
  else
    result = ch - '0';
  while (true) {
    ch = getchar();
    if (ch < '0' || ch > '9') break;
    result = result * 10 + (ch - '0');
  }
  if (minus) {
    return -result;
  } else {
    return result;
  }
}
long long fpow(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power % 2 == 1) result = (result * base);
    base = (base * base);
    power /= 2;
  }
  return result;
}
long long n, b[300000], c[300000], a[300000], d[300000], b1[300000], c1[300000];
long long bits[31][300000], kbit[31];
int main() {
  n = read();
  long long i, j, sum = 0;
  for (i = 0; i < n; i++) {
    b[i] = read();
  }
  for (i = 0; i < n; i++) {
    c[i] = read();
    d[i] = b[i] + c[i];
  }
  for (i = 0; i < n; i++) {
    sum += d[i];
  }
  sum /= 2 * n;
  for (i = 0; i < n; i++) {
    a[i] = (d[i] - sum) / n;
    if (a[i] < 0) {
      return 0 * printf("-1\n");
    }
  }
  for (i = 0; i < 31; i++) {
    for (j = 0; j < n; j++) {
      if ((a[j] & (1LL << i)) == 0)
        bits[i][j] = 0;
      else
        bits[i][j] = 1;
    }
  }
  for (i = 0; i < 31; i++) {
    kbit[i] = 0;
    for (j = 0; j < n; j++) {
      kbit[i] += bits[i][j];
    }
  }
  for (i = 0; i < n; i++) {
    b1[i] = 0;
    c1[i] = 0;
  }
  for (i = 0; i < 31; i++) {
    for (j = 0; j < n; j++) {
      int bbs = bits[i][j] ? kbit[i] : 0;
      int cbs = bits[i][j] ? n : kbit[i];
      b1[j] += bbs << i;
      c1[j] += cbs << i;
    }
  }
  for (i = 0; i < n; i++) {
    if (b1[i] != b[i] || c1[i] != c[i]) {
      return 0 * printf("-1\n");
    }
  }
  for (i = 0; i < n; i++) {
    printf("%lld ", a[i]);
  }
  printf("\n");
  return 0;
}

```