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;
struct st {
  int f, s, o;
};
vector<vector<st> > vp(4);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> ar(n - 1), br(n - 1);
  vector<vector<int> > tr;
  for (auto &i : ar) cin >> i;
  for (auto &i : br) cin >> i;
  vp[0].push_back({0, 0, 1});
  vp[0].push_back({0, 1, 0});
  vp[0].push_back({0, 2, 0});
  vp[0].push_back({0, 3, 0});
  vp[0].push_back({1, 2, 0});
  vp[1].push_back({0, 1, 1});
  vp[1].push_back({1, 1, 1});
  vp[1].push_back({1, 1, 0});
  vp[1].push_back({1, 3, 0});
  vp[2].push_back({0, 2, 1});
  vp[2].push_back({2, 2, 1});
  vp[2].push_back({2, 2, 0});
  vp[2].push_back({2, 2, 0});
  vp[2].push_back({2, 3, 0});
  vp[3].push_back({0, 3, 1});
  vp[3].push_back({1, 3, 1});
  vp[3].push_back({2, 3, 1});
  vp[3].push_back({3, 3, 1});
  vp[3].push_back({3, 3, 0});
  for (int j = 0; j < (int)vp[ar[0]].size(); ++j) {
    if (vp[ar[0]][j].o == 1 && (vp[ar[0]][j].f & vp[ar[0]][j].s) == br[0]) {
      tr.push_back({});
      tr[(int)tr.size() - 1].resize(n);
      tr[(int)tr.size() - 1][0] = (vp[ar[0]][j].f);
      tr[(int)tr.size() - 1][1] = (vp[ar[0]][j].s);
    }
  }
  for (int z = 0; z < (int)tr.size(); ++z) {
    for (int i = 1; i < n - 1; ++i) {
      for (int j = 0; j < (int)vp[ar[i]].size(); ++j) {
        int f = vp[ar[i]][j].f;
        int s = vp[ar[i]][j].s;
        int o = vp[ar[i]][j].o;
        if ((f & s) != br[i] || o != 1) continue;
        if (f == tr[z][i]) {
          tr[z][i + 1] = s;
        } else {
          tr[z][i + 1] = f;
        }
      }
    }
  }
  bool ans = 0;
  for (int z = 0; z < (int)tr.size(); ++z) {
    ans = 1;
    for (int i = 0; i < n - 1; ++i) {
      if (!((tr[z][i] | tr[z][i + 1]) == ar[i] &&
            (tr[z][i] & tr[z][i + 1]) == br[i]))
        ans = 0;
    }
    if (ans) {
      cout << "YES\n";
      for (int i = 0; i < n; ++i) cout << tr[z][i] << " ";
      cout << "\n";
      return 0;
    }
  }
  cout << "NO\n";
  if (false)
    cerr << "Time execute: " << clock() / (double)CLOCKS_PER_SEC << " sec"
         << endl;
}
