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>
const int N = (int)1e5 + 1, INF = (int)1e9;
const float PI = 3.14159265359, EPS = 1e-6;
using namespace std;
string to_base(int n, int p) {
  int r;
  string s;
  while (n) {
    r = n % p;
    n /= p;
    s += (r + '0');
  }
  reverse(s.begin(), s.end());
  while (s.size() < 2) s.insert(s.begin(), '0');
  return s;
}
string s, t[100001], tmp, v, b;
int k, n, w, h, ans, x, p;
bool f;
char c;
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> x;
    s += (x + '0');
  }
  for (int i = 0; i < n - 1; i++) {
    cin >> x;
    v += (x + '0');
  }
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == '0' && v[i] != '0' ||
        s[i] == '1' && (v[i] == '2' || v[i] == '3') ||
        s[i] == '2' && v[i] == '1' || s[i] == '2' && v[i] == '3')
      f = 1;
  }
  if (!f) {
    int j = 0;
    while (j < n) {
      if (s[j] == v[j]) {
        t[k].push_back(s[j]);
        t[k].push_back(s[j]);
        k++;
      } else if (s[j] == '3' && v[j] == '0') {
        t[k].push_back('2');
        t[k].push_back('1');
        t[k].push_back('1');
        t[k].push_back('2');
        t[k].push_back('3');
        t[k].push_back('0');
        t[k].push_back('0');
        t[k].push_back('3');
        k++;
      } else {
        t[k].push_back(s[j]);
        t[k].push_back(v[j]);
        t[k].push_back(v[j]);
        t[k].push_back(s[j]);
        k++;
      }
      j++;
    }
    k--;
    if (k == 1) {
      b.push_back(t[0][0]), b.push_back(t[0][1]);
    } else {
      for (int i = 0; i < k - 1; i++) {
        for (int j = 1; j < t[i].size(); j += 2) {
          for (int r = 0; r < t[i + 1].size(); r += 2) {
            if (i < 2 || i == k - 2) {
              if (t[i][j] == t[i + 1][r]) {
                c = t[i + 1][r + 1];
                if (i == 0) b.push_back(t[i][j - 1]);
                b.push_back(t[i][j]);
                if (i == k - 2) b.push_back(t[i + 1][r + 1]);
                p++;
              }
            } else {
              if (t[i][j] == t[i + 1][r] && t[i + 1][r] == c) {
                c = t[i + 1][r + 1];
                b.push_back(t[i][j]);
                p++;
              }
            }
          }
          if (p > 0) goto kek;
        }
      kek:
        if (p == 0) f = 1;
        p = 0;
      }
    }
    if (f)
      cout << "NO";
    else {
      cout << "YES" << endl;
      for (int i = 0; i < b.size(); i++) cout << b[i] << " ";
    }
  } else
    cout << "NO";
  return 0;
}
