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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
inline void Routine() {
  srand(time(NULL));
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &p) {
  in >> p.first >> p.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, const pair<T, U> &p) {
  out << p.first << ' ' << p.second;
  return out;
}
template <class T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &i : v) {
    in >> i;
  }
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (auto &i : v) {
    out << i << ' ';
  }
  return out;
}
int main() {
  Routine();
  long long b, c, d, n, ans = 0;
  string s, s1;
  cin >> n;
  int a[n];
  int m[n];
  int an[n];
  for (int i = 0; i < n - 1; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    cin >> m[i];
    if (m[i] > a[i]) {
      cout << "NO";
      return 0;
    }
  }
  an[0] = m[0];
  for (int i = 1; i < n - 1; i++) {
    an[i] = m[i - 1] | m[i];
  }
  an[n - 1] = m[n - 2];
  for (int i = 0; i < n - 1; i++) {
    if (((an[i]) | (an[i + 1])) != a[i]) {
      if (i < n - 2) {
        long long k = a[i] ^ a[i + 1];
        long long ki = k & a[i];
        an[i] = an[i] | ki;
        an[i + 2] = an[i + 2] | (k & a[i + 1]);
        if (((an[i]) | (an[i + 1])) != a[i]) {
          an[i + 1] = (a[i] & (a[i] ^ an[i])) | an[i + 1];
        }
      } else {
        an[i + 1] = (a[i] & (a[i] ^ an[i])) | an[i + 1];
      }
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (((an[i]) | (an[i + 1])) != a[i]) {
      cout << "NO";
      return 0;
    }
    if (((an[i]) & (an[i + 1])) != m[i]) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cout << an[i] << " ";
  }
  return 0;
}
