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;
const int md = 1e9 + 7;
const int MAX = 1e5 + 5;
const int MAXN = 1e6 + 5;
map<pair<long long, long long>, pair<long long, long long> > mp;
long long n;
long long a[MAX], b[MAX];
bool done = false;
void fun(vector<long long> ans, long long ind) {
  if (ind == n - 1) {
    done = true;
    cout << "YES\n";
    for (long long i = 0; i < ans.size(); i += 1) cout << ans[i] << " ";
    exit(0);
  }
  if (ind == 0) {
    ans.push_back(mp[{a[ind], b[ind]}].first);
    ans.push_back(mp[{a[ind], b[ind]}].second);
    fun(ans, ind + 1);
    ans.clear();
    ans.push_back(mp[{a[ind], b[ind]}].second);
    ans.push_back(mp[{a[ind], b[ind]}].first);
    fun(ans, ind + 1);
  } else {
    long long last = ans.size() - 1;
    long long t2 = ans[last];
    if (mp[{a[ind], b[ind]}].first == t2) {
      ans.push_back(mp[{a[ind], b[ind]}].second);
      fun(ans, ind + 1);
    } else if (mp[{a[ind], b[ind]}].second == t2) {
      ans.push_back(mp[{a[ind], b[ind]}].first);
      fun(ans, ind + 1);
    } else {
      return;
    }
  }
  return;
}
void solve() {
  vector<pair<pair<long long, long long>, pair<long long, long long> > > v;
  for (long long i = 0; i < 4; i += 1) {
    for (long long j = i; j < 4; j += 1) {
      long long OR = i | j;
      long long AND = i & j;
      v.push_back({{i, j}, {OR, AND}});
    }
  }
  for (long long i = 0; i < v.size(); i += 1) {
    mp[{v[i].second.first, v[i].second.second}] =
        make_pair(v[i].first.first, v[i].first.second);
  }
  cin >> n;
  for (long long i = 0; i < n - 1; i += 1) cin >> a[i];
  for (long long i = 0; i < n - 1; i += 1) cin >> b[i];
  vector<long long> ans;
  fun(ans, 0);
  if (!done) cout << "NO";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}
