#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100000;
char a[MAX_N][2], b[MAX_N][2];
int t[MAX_N][2];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, tmp;
  cin >> n;
  for (int i = 0; i < n - 1; ++i)
    cin >> tmp, a[i][0] = tmp & 1, a[i][1] = (tmp >> 1) & 1;
  for (int i = 0; i < n - 1; ++i)
    cin >> tmp, b[i][0] = tmp & 1, b[i][1] = (tmp >> 1) & 1;
  for (int i = 0; i < n - 1; ++i) {
    for (int j = 0; j < 2; ++j) {
      if (a[i][j] == 0 && b[i][j] == 0) {
        if (t[i][j] == 2) return cout << "NO\n", 0;
        t[i][j] = 1;
        t[i + 1][j] = 1;
      } else if (a[i][j] == 1 && b[i][j] == 0) {
        if (t[i][j] == 0)
          t[i][j] = 3;
        else
          t[i + 1][j] = 3 - t[i][j];
      } else if (a[i][j] == 1 && b[i][j] == 1) {
        if (t[i][j] == 1) return cout << "NO\n", 0;
        t[i][j] = 2;
        t[i + 1][j] = 2;
      } else {
        return cout << "NO\n", 0;
      }
    }
  }
  for (int j = 0; j < 2; ++j) {
    int firstOk = -1;
    for (int i = 0; i < n - 1; ++i) {
      if (t[i][j] != 3) {
        firstOk = i;
        break;
      }
    }
    if (firstOk == -1)
      for (int i = 0; i < n; ++i) t[i][j] = 1 + (i & 1);
    else {
      for (int i = firstOk; i--;) t[i][j] = 3 - t[i + 1][j];
      for (int i = firstOk + 1; i < n; ++i)
        if (t[i][j] == 3) {
          if (i != n - 2 && t[i + 1][j] != 3 && 3 - t[i + 1][j] == t[i][j])
            return cout << "NO\n", 0;
          else
            t[i][j] = 3 - t[i - 1][j];
        }
    }
  }
  cout << "YES\n";
  for (int i = 0; i < n; ++i) cout << ((t[i][1] - 1) * 2) + t[i][0] - 1 << " ";
  cout << '\n';
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const int MAX_N = 100000;
char a[MAX_N][2], b[MAX_N][2];
int t[MAX_N][2];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, tmp;
  cin >> n;
  for (int i = 0; i < n - 1; ++i)
    cin >> tmp, a[i][0] = tmp & 1, a[i][1] = (tmp >> 1) & 1;
  for (int i = 0; i < n - 1; ++i)
    cin >> tmp, b[i][0] = tmp & 1, b[i][1] = (tmp >> 1) & 1;
  for (int i = 0; i < n - 1; ++i) {
    for (int j = 0; j < 2; ++j) {
      if (a[i][j] == 0 && b[i][j] == 0) {
        if (t[i][j] == 2) return cout << "NO\n", 0;
        t[i][j] = 1;
        t[i + 1][j] = 1;
      } else if (a[i][j] == 1 && b[i][j] == 0) {
        if (t[i][j] == 0)
          t[i][j] = 3;
        else
          t[i + 1][j] = 3 - t[i][j];
      } else if (a[i][j] == 1 && b[i][j] == 1) {
        if (t[i][j] == 1) return cout << "NO\n", 0;
        t[i][j] = 2;
        t[i + 1][j] = 2;
      } else {
        return cout << "NO\n", 0;
      }
    }
  }
  for (int j = 0; j < 2; ++j) {
    int firstOk = -1;
    for (int i = 0; i < n - 1; ++i) {
      if (t[i][j] != 3) {
        firstOk = i;
        break;
      }
    }
    if (firstOk == -1)
      for (int i = 0; i < n; ++i) t[i][j] = 1 + (i & 1);
    else {
      for (int i = firstOk; i--;) t[i][j] = 3 - t[i + 1][j];
      for (int i = firstOk + 1; i < n; ++i)
        if (t[i][j] == 3) {
          if (i != n - 2 && t[i + 1][j] != 3 && 3 - t[i + 1][j] == t[i][j])
            return cout << "NO\n", 0;
          else
            t[i][j] = 3 - t[i - 1][j];
        }
    }
  }
  cout << "YES\n";
  for (int i = 0; i < n; ++i) cout << ((t[i][1] - 1) * 2) + t[i][0] - 1 << " ";
  cout << '\n';
  return 0;
}

```