#include <bits/stdc++.h>
using namespace std;
int n;
string s;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  vector<vector<int>> Res(n, vector<int>(2, 0)), A(n - 1, vector<int>(2, 0)),
      B(n - 1, vector<int>(2, 0));
  for (int i = 0; i < n - 1; i++) {
    int num;
    cin >> num;
    if (num == 3) {
      A[i][1] = 1;
      A[i][0] = 1;
    }
    if (num == 2) {
      A[i][1] = 1;
      A[i][0] = 0;
    }
    if (num == 1) {
      A[i][1] = 0;
      A[i][0] = 1;
    }
    if (num == 0) {
      A[i][1] = 0;
      A[i][0] = 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int num;
    cin >> num;
    if (num == 3) {
      B[i][1] = 1;
      B[i][0] = 1;
    }
    if (num == 2) {
      B[i][1] = 1;
      B[i][0] = 0;
    }
    if (num == 1) {
      B[i][1] = 0;
      B[i][0] = 1;
    }
    if (num == 0) {
      B[i][1] = 0;
      B[i][0] = 0;
    }
  }
  for (int t = 0; t <= 3; t++) {
    bool ok = 1;
    if (t == 3) {
      Res[0][1] = 1;
      Res[0][0] = 1;
    }
    if (t == 2) {
      Res[0][1] = 1;
      Res[0][0] = 0;
    }
    if (t == 1) {
      Res[0][1] = 0;
      Res[0][0] = 1;
    }
    if (t == 0) {
      Res[0][1] = 0;
      Res[0][0] = 0;
    }
    for (int i = 1; i < n; i++) {
      if (!ok) break;
      for (int j = 0; j <= 1; j++) {
        if (A[i - 1][j] == 1 && B[i - 1][j] == 1) {
          if (Res[i - 1][j] == 0) {
            ok = 0;
            break;
          };
          Res[i][j] = 1;
        }
        if (A[i - 1][j] == 1 && B[i - 1][j] == 0) {
          if (Res[i - 1][j] == 0) {
            Res[i][j] = 1;
          };
          if (Res[i - 1][j] == 1) {
            Res[i][j] = 0;
          };
        }
        if (A[i - 1][j] == 0 && B[i - 1][j] == 1) {
          ok = 0;
          break;
        }
        if (A[i - 1][j] == 0 && B[i - 1][j] == 0) {
          if (Res[i - 1][j] == 1) {
            ok = 0;
            break;
          };
          Res[i][j] = 0;
        }
      }
    }
    if (ok) {
      cout << "YES\n";
      for (int i = 0; i < n; i++) {
        cout << Res[i][1] * 2 + Res[i][0] << ' ';
      }
      return 0;
    }
  }
  cout << "NO\n";
}

### 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;
int n;
string s;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  vector<vector<int>> Res(n, vector<int>(2, 0)), A(n - 1, vector<int>(2, 0)),
      B(n - 1, vector<int>(2, 0));
  for (int i = 0; i < n - 1; i++) {
    int num;
    cin >> num;
    if (num == 3) {
      A[i][1] = 1;
      A[i][0] = 1;
    }
    if (num == 2) {
      A[i][1] = 1;
      A[i][0] = 0;
    }
    if (num == 1) {
      A[i][1] = 0;
      A[i][0] = 1;
    }
    if (num == 0) {
      A[i][1] = 0;
      A[i][0] = 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int num;
    cin >> num;
    if (num == 3) {
      B[i][1] = 1;
      B[i][0] = 1;
    }
    if (num == 2) {
      B[i][1] = 1;
      B[i][0] = 0;
    }
    if (num == 1) {
      B[i][1] = 0;
      B[i][0] = 1;
    }
    if (num == 0) {
      B[i][1] = 0;
      B[i][0] = 0;
    }
  }
  for (int t = 0; t <= 3; t++) {
    bool ok = 1;
    if (t == 3) {
      Res[0][1] = 1;
      Res[0][0] = 1;
    }
    if (t == 2) {
      Res[0][1] = 1;
      Res[0][0] = 0;
    }
    if (t == 1) {
      Res[0][1] = 0;
      Res[0][0] = 1;
    }
    if (t == 0) {
      Res[0][1] = 0;
      Res[0][0] = 0;
    }
    for (int i = 1; i < n; i++) {
      if (!ok) break;
      for (int j = 0; j <= 1; j++) {
        if (A[i - 1][j] == 1 && B[i - 1][j] == 1) {
          if (Res[i - 1][j] == 0) {
            ok = 0;
            break;
          };
          Res[i][j] = 1;
        }
        if (A[i - 1][j] == 1 && B[i - 1][j] == 0) {
          if (Res[i - 1][j] == 0) {
            Res[i][j] = 1;
          };
          if (Res[i - 1][j] == 1) {
            Res[i][j] = 0;
          };
        }
        if (A[i - 1][j] == 0 && B[i - 1][j] == 1) {
          ok = 0;
          break;
        }
        if (A[i - 1][j] == 0 && B[i - 1][j] == 0) {
          if (Res[i - 1][j] == 1) {
            ok = 0;
            break;
          };
          Res[i][j] = 0;
        }
      }
    }
    if (ok) {
      cout << "YES\n";
      for (int i = 0; i < n; i++) {
        cout << Res[i][1] * 2 + Res[i][0] << ' ';
      }
      return 0;
    }
  }
  cout << "NO\n";
}

```