#include <bits/stdc++.h>
using namespace std;
const int N = 100007;
int t[N], a[N], b[N], n;
bool dfs(int i) {
  if (i + 1 == n) return true;
  for (int j = 0; j < 4; j++) {
    if ((j | t[i]) == a[i] && (j & t[i]) == b[i]) {
      t[i + 1] = j;
      if (dfs(i + 1)) return true;
    }
  }
  return false;
}
void print_vec(int n) {
  if (n == 0) cout << endl;
  cout << t[0];
  for (int i = 1; i < n; i++) cout << ' ' << t[i];
  cout << endl;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d", &b[i]);
  }
  bool f = 1;
  for (int i = 0; i < 4; i++) {
    t[0] = i;
    if (dfs(0)) {
      cout << "YES" << endl;
      print_vec(n);
      f = 0;
      break;
    }
  }
  if (f) cout << "NO" << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int N = 100007;
int t[N], a[N], b[N], n;
bool dfs(int i) {
  if (i + 1 == n) return true;
  for (int j = 0; j < 4; j++) {
    if ((j | t[i]) == a[i] && (j & t[i]) == b[i]) {
      t[i + 1] = j;
      if (dfs(i + 1)) return true;
    }
  }
  return false;
}
void print_vec(int n) {
  if (n == 0) cout << endl;
  cout << t[0];
  for (int i = 1; i < n; i++) cout << ' ' << t[i];
  cout << endl;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d", &b[i]);
  }
  bool f = 1;
  for (int i = 0; i < 4; i++) {
    t[0] = i;
    if (dfs(0)) {
      cout << "YES" << endl;
      print_vec(n);
      f = 0;
      break;
    }
  }
  if (f) cout << "NO" << endl;
}

```