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<pair<int, int> > dp[20];
bool solve(int val, int idx, int a[], int b[], int t[], int d[][4], int n) {
  if (idx == n + 1) {
    return true;
  }
  if (d[idx][val] != -1) return false;
  int size = dp[4 * a[idx] + b[idx]].size();
  for (int i = 0; i < size; i++) {
    pair<int, int> p = dp[4 * a[idx] + b[idx]][i];
    if (val == p.first) {
      t[idx + 1] = p.second;
      bool flag = solve(p.second, idx + 1, a, b, t, d, n);
      if (flag) return true;
    }
  }
  d[idx][val] = 0;
  return false;
}
int main() {
  int n;
  cin >> n;
  dp[0].push_back(make_pair(0, 0));
  dp[1].push_back(make_pair(0, 1));
  dp[1].push_back(make_pair(1, 0));
  dp[2].push_back(make_pair(0, 2));
  dp[2].push_back(make_pair(2, 0));
  dp[3].push_back(make_pair(1, 2));
  dp[3].push_back(make_pair(2, 1));
  dp[3].push_back(make_pair(0, 3));
  dp[3].push_back(make_pair(3, 0));
  dp[5].push_back(make_pair(1, 1));
  dp[7].push_back(make_pair(1, 3));
  dp[7].push_back(make_pair(3, 1));
  dp[10].push_back(make_pair(2, 2));
  dp[11].push_back(make_pair(2, 3));
  dp[11].push_back(make_pair(3, 2));
  dp[15].push_back(make_pair(3, 3));
  int a[200002], b[200005];
  int d[200004][4];
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= 3; j++) {
      d[i][j] = -1;
    }
  }
  for (int i = 1; i < n; i++) {
    cin >> b[i];
  }
  for (int i = 1; i < n; i++) {
    cin >> a[i];
  }
  int t[n + 1];
  bool flag = false;
  for (int j = 0; j < dp[4 * a[1] + b[1]].size(); j++) {
    pair<int, int> p = dp[4 * a[1] + b[1]][j];
    t[1] = p.first;
    t[2] = p.second;
    flag = false;
    if (solve(p.second, 2, a, b, t, d, n)) {
      cout << "YES" << endl;
      for (int k = 1; k <= n; k++) {
        cout << t[k] << " ";
      }
      cout << endl;
      flag = true;
      break;
    }
  }
  if (flag == 0) cout << "NO" << endl;
  return 0;
}
