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;
int main() {
  int n;
  cin >> n;
  int a[n - 1], b[n - 1];
  for (int i = 0; i < n - 1; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) cin >> b[i];
  map<pair<int, int>, pair<int, int> > m;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) m[{i, j}] = {-1, -1};
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      for (int ii = 0; ii < 2; ii++)
        for (int jj = 0; jj < 2; jj++)
          if (((ii | jj) == i) && ((ii & jj) == j)) m[{i, j}] = {ii, jj};
  bool ans = 1;
  for (int i = 0; i < n - 1; i++)
    if (m[{a[i], b[i]}].first == -1) ans = 0;
  if (!ans) {
    cout << "NO";
    return 0;
  }
  int t[n], t2[n];
  {
    t[0] = m[{a[0] & 1, b[0] & 1}].first;
    t[1] = m[{a[0] & 1, b[0] & 1}].second;
    for (int i = 2; i < n; i++) {
      pair<int, int> p = m[{a[i - 1] & 1, b[i - 1] & 1}];
      bool B = 0;
      if (p.first == t[i - 1]) {
        t[i] = p.second;
        B = 1;
      }
      if (p.second == t[i - 1]) {
        t[i] = p.first;
        B = 1;
      }
      ans &= B;
    }
    if (!ans) {
      ans = 1;
      t[0] = m[{a[0] & 1, b[0] & 1}].second;
      t[1] = m[{a[0] & 1, b[0] & 1}].first;
      for (int i = 2; i < n; i++) {
        pair<int, int> p = m[{a[i - 1] & 1, b[i - 1] & 1}];
        bool B = 0;
        if (p.first == t[i - 1]) {
          t[i] = p.second;
          B = 1;
        }
        if (p.second == t[i - 1]) {
          t[i] = p.first;
          B = 1;
        }
        ans &= B;
      }
      if (!ans) {
        cout << "NO";
        return 0;
      }
    }
  }
  for (int i = 0; i < n - 1; i++) {
    a[i] /= 2;
    b[i] /= 2;
  }
  {
    t2[0] = m[{a[0] & 1, b[0] & 1}].first;
    t2[1] = m[{a[0] & 1, b[0] & 1}].second;
    for (int i = 2; i < n; i++) {
      pair<int, int> p = m[{a[i - 1] & 1, b[i - 1] & 1}];
      bool B = 0;
      if (p.first == t2[i - 1]) {
        t2[i] = p.second;
        B = 1;
      }
      if (p.second == t2[i - 1]) {
        t2[i] = p.first;
        B = 1;
      }
      ans &= B;
    }
    if (!ans) {
      ans = 1;
      t2[0] = m[{a[0] & 1, b[0] & 1}].second;
      t2[1] = m[{a[0] & 1, b[0] & 1}].first;
      for (int i = 2; i < n; i++) {
        pair<int, int> p = m[{a[i - 1] & 1, b[i - 1] & 1}];
        bool B = 0;
        if (p.first == t2[i - 1]) {
          t2[i] = p.second;
          B = 1;
        }
        if (p.second == t2[i - 1]) {
          t2[i] = p.first;
          B = 1;
        }
        ans &= B;
      }
      if (!ans) {
        cout << "NO";
        return 0;
      }
    }
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) cout << t[i] + 2 * t2[i] << " ";
  return 0;
}
