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.
#include <bits/stdc++.h>
using namespace std;
vector<int> solve(vector<int> &a, vector<int> &b, int n) {
  vector<vector<int>> d(n + 1, vector<int>(2, 0)), p(n + 1, vector<int>(2, -1));
  vector<int> out;
  d[0][0] = d[0][1] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++)
        if (d[i - 1][j] == 1 && (j | k) == a[i - 1] && b[i - 1] == (j & k)) {
          d[i][k] = 1;
          p[i][k] = j;
        }
  }
  int y = 0;
  if (!d[n][y] && !d[n][y + 1])
    return out;
  else if (d[n][y + 1])
    y++;
  for (int i = n; i >= 0; i--) {
    out.emplace_back(y);
    y = p[i][y];
  }
  reverse(out.begin(), out.end());
  return out;
}
int main() {
  int n;
  cin >> n;
  n--;
  vector<int> a1(n), a2(n), b1(n), b2(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a1[i] = ((x & 2) >> 1);
    a2[i] = (x & 1);
  }
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    b1[i] = ((x & 2) >> 1);
    b2[i] = (x & 1);
  }
  vector<int> OTBET1 = solve(a1, b1, n), OTBET2 = solve(a2, b2, n);
  if (OTBET1.size() != n + 1 || OTBET2.size() != n + 1)
    return cout << "NO\n", 0;
  cout << "YES\n";
  for (int i = 0; i <= n; i++) {
    cout << OTBET1[i] * 2 + OTBET2[i] << ' ';
  }
  return 0;
}
