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;
void Debug() { cout << '\n'; }
template <class FIRST, class... REST>
void Debug(FIRST arg, REST... rest) {
  cout << arg << " ";
  Debug(rest...);
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  out << "[";
  if (!v.empty()) {
    for (int i = int(0); i < int((int)(v).size() - 1); i++) out << v[i] << ", ";
    out << v.back();
  }
  out << "]";
  return out;
}
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& v) {
  out << "(" << v.first << ", " << v.second << ")";
  return out;
}
const int MAX_N = 300010;
const int MAX_V = 100010;
const double eps = 1e-6;
const long long mod = 1000000007;
const int inf = 1 << 30;
const long long linf = 1LL << 60;
const double PI = 3.14159265358979323846;
int N;
int A[MAX_N], B[MAX_N];
int ans[MAX_N];
void solve() {
  cin >> N;
  for (int i = int(0); i < int(N - 1); i++) cin >> A[i];
  for (int i = int(0); i < int(N - 1); i++) cin >> B[i];
  for (int q = int(0); q < int(4); q++) {
    ans[0] = q;
    bool ok = true;
    for (int i = int(0); i < int(N - 1); i++) {
      bool found = false;
      for (int j = int(0); j < int(4); j++) {
        if ((ans[i] | j) == A[i] && (ans[i] & j) == B[i]) {
          found = true;
          ans[i + 1] = j;
          break;
        }
      }
      if (!found) {
        ok = false;
        break;
      }
    }
    if (ok) {
      cout << "YES\n";
      for (int i = int(0); i < int(N); i++) {
        cout << ans[i] << " ";
      }
      cout << "\n";
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed;
  cout.precision(20);
  srand((unsigned int)time(NULL));
  solve();
  return 0;
}
