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 usefiles(const string &s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
template <class t1, class t2>
inline bool cmin(t1 &a, const t2 &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class t1, class t2>
inline bool cmax(t1 &a, const t2 &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
void run();
signed main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  run();
}
long long mex(vector<long long> v) {
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (long long i = 0; i < ((long long)v.size()); ++i)
    if (i != v[i]) return i;
  return ((long long)v.size());
}
void run() {
  long long n;
  cin >> n;
  vector<long long> a(n - 1), b(n - 1);
  for (auto &x : a) cin >> x;
  for (auto &x : b) cin >> x;
  vector<vector<long long>> dp(n, vector<long long>(4, 0));
  for (long long i = 0; i < 4; ++i) dp[0][i] = 1;
  vector<vector<long long>> prv(n, vector<long long>(4, -1));
  for (long long i = 1; i < n; ++i) {
    for (long long j = 0; j < 4; ++j)
      for (long long k = 0; k < 4; ++k)
        if ((j | k) == a[i - 1] && (j & k) == b[i - 1] &&
            cmax(dp[i][j], dp[i - 1][k]))
          prv[i][j] = k;
  }
  for (long long i = 0; i < 4; ++i)
    if (dp.back()[i]) {
      cout << "YES" << endl;
      long long cur = i;
      vector<long long> res;
      for (long long j = n - 1; j > -1; --j) {
        res.push_back(cur);
        cur = prv[j][cur];
      }
      reverse(res.begin(), res.end());
      for (auto &x : res) cout << x << " ";
      cout << endl;
      return;
    }
  cout << "NO" << endl;
}
