#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void DEBUG(string label, T value) {
  cerr << "[" << label << " = " << value << "]\n";
}
int main(int argv, char **argc) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int g[4][4][4];
  for (int i = 0; i <= 3; ++i) {
    for (int j = 0; j <= 3; ++j) {
      for (int k = 0; k <= 3; ++k) {
        g[i][j][k] = -1;
      }
    }
  }
  for (int a = 0; a <= 3; ++a) {
    for (int b = 0; b <= 3; ++b) {
      for (int c = 0; c <= 3; ++c) {
        for (int d = 0; d <= 3; ++d) {
          if ((c | d) == a && (c & d) == b) {
            g[a][b][c] = d;
            break;
          }
        }
      }
    }
  }
  int n;
  cin >> n;
  int a[n], b[n];
  for (int i = 0; i < n - 1; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; ++i) {
    cin >> b[i];
  }
  for (int i = 0; i <= 3; ++i) {
    vector<int> ans(n);
    ans[0] = i;
    bool found = true;
    for (int j = 1; j < n; ++j) {
      if (g[a[j - 1]][b[j - 1]][ans[j - 1]] == -1) {
        found = false;
        break;
      }
      ans[j] = g[a[j - 1]][b[j - 1]][ans[j - 1]];
    }
    if (found) {
      cout << "YES\n";
      for (auto e : ans) {
        cout << e << " ";
      }
      cout << '\n';
      return 0;
    }
  }
  cout << "NO\n";
}

### Prompt

Please formulate 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void DEBUG(string label, T value) {
  cerr << "[" << label << " = " << value << "]\n";
}
int main(int argv, char **argc) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int g[4][4][4];
  for (int i = 0; i <= 3; ++i) {
    for (int j = 0; j <= 3; ++j) {
      for (int k = 0; k <= 3; ++k) {
        g[i][j][k] = -1;
      }
    }
  }
  for (int a = 0; a <= 3; ++a) {
    for (int b = 0; b <= 3; ++b) {
      for (int c = 0; c <= 3; ++c) {
        for (int d = 0; d <= 3; ++d) {
          if ((c | d) == a && (c & d) == b) {
            g[a][b][c] = d;
            break;
          }
        }
      }
    }
  }
  int n;
  cin >> n;
  int a[n], b[n];
  for (int i = 0; i < n - 1; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; ++i) {
    cin >> b[i];
  }
  for (int i = 0; i <= 3; ++i) {
    vector<int> ans(n);
    ans[0] = i;
    bool found = true;
    for (int j = 1; j < n; ++j) {
      if (g[a[j - 1]][b[j - 1]][ans[j - 1]] == -1) {
        found = false;
        break;
      }
      ans[j] = g[a[j - 1]][b[j - 1]][ans[j - 1]];
    }
    if (found) {
      cout << "YES\n";
      for (auto e : ans) {
        cout << e << " ";
      }
      cout << '\n';
      return 0;
    }
  }
  cout << "NO\n";
}

```