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;
using ll = long long;
using ull = unsigned long long;
vector<set<pair<int, int>>> can_get_by_or(4);
vector<set<pair<int, int>>> can_get_by_and(4);
vector<pair<int, int>> pr;
vector<int> ans;
bool dfs(int i, bool first) {
  if (i == pr.size() - 1) {
    if (first)
      ans.push_back(pr[i].first);
    else
      ans.push_back(pr[i].second);
    return true;
  }
  if (first) {
    if (pr[i].first == pr[i + 1].first) {
      if (dfs(i + 1, false)) {
        ans.push_back(pr[i].first);
        return true;
      }
    }
    if (pr[i].first == pr[i + 1].second) {
      if (dfs(i + 1, true)) {
        ans.push_back(pr[i].first);
        return true;
      }
    }
  } else {
    if (pr[i].second == pr[i + 1].first) {
      if (dfs(i + 1, false)) {
        ans.push_back(pr[i].second);
        return true;
      }
    }
    if (pr[i].second == pr[i + 1].second) {
      if (dfs(i + 1, true)) {
        ans.push_back(pr[i].second);
        return true;
      }
    }
  }
  return false;
}
int solve(istream& cin, ostream& cout) {
  for (int i = 0; i < 4; i++) {
    for (int a = 0; a < 4; a++) {
      for (int b = 0; b < 4; b++) {
        if ((a & b) == i) {
          can_get_by_and[i].insert({a, b});
        }
        if ((a | b) == i) {
          can_get_by_or[i].insert({a, b});
        }
      }
    }
  }
  int n;
  cin >> n;
  vector<int> a(n - 1), b(n - 1);
  for (int& i : a) {
    cin >> i;
  }
  for (int& i : b) {
    cin >> i;
  }
  pr.resize(n - 1);
  for (int i = 0; i < n - 1; i++) {
    ;
    ;
    for (auto p : can_get_by_or[a[i]]) {
      if (can_get_by_and[b[i]].count(p)) {
        pr[i] = p;
        ;
        ;
        break;
      }
    }
  }
  ans.reserve(n);
  if (dfs(0, false)) {
    cout << "YES\n";
    cout << pr[0].first << ' ';
    for (auto it = ans.rbegin(); it != ans.rend(); it++) {
      cout << *it << ' ';
    }
  } else if (dfs(0, true)) {
    cout << "YES\n";
    cout << pr[0].second << ' ';
    for (auto it = ans.rbegin(); it != ans.rend(); it++) {
      cout << *it << ' ';
    }
  } else {
    cout << "NO\n";
  }
}
void test() {
  vector<pair<string, string>> tests = {};
  for (auto [in, out] : tests) {
    stringstream input(in);
    stringstream output;
    solve(input, output);
    if (output.str() != out) {
      cout << "test failed. Input:\n " << in << "\nRight answer:\n"
           << out << "\nasnwer:\n"
           << output.str() << endl;
      exit(1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve(cin, cout);
  return 0;
}
