#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int main() {
  int n = read();
  int a[n - 1];
  int b[n - 1];
  int t[n];
  for (register int i = (int)(0); i < (int)(n - 1); i++) a[i] = read();
  for (register int i = (int)(0); i < (int)(n - 1); i++) b[i] = read();
  int flag_out = 0;
  for (register int i = (int)(0); i < (int)(4); i++) {
    t[0] = i;
    for (register int ii = (int)(1); ii < (int)(n); ii++) {
      int flag = 0;
      for (register int iii = (int)(0); iii < (int)(4); iii++) {
        if (((iii | t[ii - 1]) == a[ii - 1]) &&
            ((iii & t[ii - 1]) == b[ii - 1])) {
          t[ii] = iii;
          flag = 1;
          break;
        }
      }
      if (!flag) goto outer;
    }
    flag_out = 1;
    break;
  outer:;
  }
  if (!flag_out) {
    printf("NO");
    exit(0);
  }
  printf("YES\n");
  for (register int i = (int)(0); i < (int)(n - 1); i++) {
    printf("%d ", t[i]);
  }
  printf("%d", t[n - 1]);
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

When Masha came to math classes today, she saw two integer sequences of length n - 1 on the blackboard. Let's denote the elements of the first sequence as a_i (0 ≤ a_i ≤ 3), and the elements of the second sequence as b_i (0 ≤ b_i ≤ 3).

Masha became interested if or not there is an integer sequence of length n, which elements we will denote as t_i (0 ≤ t_i ≤ 3), so that for every i (1 ≤ i ≤ n - 1) the following is true: 

  * a_i = t_i | t_{i + 1} (where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#OR)) and 
  * b_i = t_i \& t_{i + 1} (where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND)). 



The question appeared to be too difficult for Masha, so now she asked you to check whether such a sequence t_i of length n exists. If it exists, find such a sequence. If there are multiple such sequences, find any of them.

Input

The first line contains a single integer n (2 ≤ n ≤ 10^5) — the length of the sequence t_i. 

The second line contains n - 1 integers a_1, a_2, …, a_{n-1} (0 ≤ a_i ≤ 3) — the first sequence on the blackboard.

The third line contains n - 1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 3) — the second sequence on the blackboard.

Output

In the first line print "YES" (without quotes), if there is a sequence t_i that satisfies the conditions from the statements, and "NO" (without quotes), if there is no such sequence.

If there is such a sequence, on the second line print n integers t_1, t_2, …, t_n (0 ≤ t_i ≤ 3) — the sequence that satisfies the statements conditions.

If there are multiple answers, print any of them.

Examples

Input

4
3 3 2
1 2 0


Output

YES
1 3 2 0 

Input

3
1 3
3 2


Output

NO

Note

In the first example it's easy to see that the sequence from output satisfies the given conditions: 

  * t_1 | t_2 = (01_2) | (11_2) = (11_2) = 3 = a_1 and t_1 \& t_2 = (01_2) \& (11_2) = (01_2) = 1 = b_1; 
  * t_2 | t_3 = (11_2) | (10_2) = (11_2) = 3 = a_2 and t_2 \& t_3 = (11_2) \& (10_2) = (10_2) = 2 = b_2; 
  * t_3 | t_4 = (10_2) | (00_2) = (10_2) = 2 = a_3 and t_3 \& t_4 = (10_2) \& (00_2) = (00_2) = 0 = b_3. 



In the second example there is no such sequence.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int main() {
  int n = read();
  int a[n - 1];
  int b[n - 1];
  int t[n];
  for (register int i = (int)(0); i < (int)(n - 1); i++) a[i] = read();
  for (register int i = (int)(0); i < (int)(n - 1); i++) b[i] = read();
  int flag_out = 0;
  for (register int i = (int)(0); i < (int)(4); i++) {
    t[0] = i;
    for (register int ii = (int)(1); ii < (int)(n); ii++) {
      int flag = 0;
      for (register int iii = (int)(0); iii < (int)(4); iii++) {
        if (((iii | t[ii - 1]) == a[ii - 1]) &&
            ((iii & t[ii - 1]) == b[ii - 1])) {
          t[ii] = iii;
          flag = 1;
          break;
        }
      }
      if (!flag) goto outer;
    }
    flag_out = 1;
    break;
  outer:;
  }
  if (!flag_out) {
    printf("NO");
    exit(0);
  }
  printf("YES\n");
  for (register int i = (int)(0); i < (int)(n - 1); i++) {
    printf("%d ", t[i]);
  }
  printf("%d", t[n - 1]);
  return 0;
}

```