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;
bool fun(int* arr, int* brr, pair<int, int>* crr, int i) {
  int a0 = arr[i] & 1, a1 = (arr[i] / 2) & 1, b0 = brr[i] & 1,
      b1 = (brr[i] / 2) & 1;
  if (a0 == 1 && b0 == 0) {
    if (crr[i].first == -1)
      crr[i + 1].first = -1;
    else
      crr[i + 1].first = 1 ^ crr[i].first;
  } else {
    if (a0 == 1 && b0 == 1) {
      if (crr[i].first == 0)
        return false;
      else
        crr[i].first = crr[i + 1].first = 1;
    } else {
      if (a0 == 0 && b0 == 1)
        return false;
      else {
        if (crr[i].first == 1)
          return false;
        else
          crr[i].first = crr[i + 1].first = 0;
      }
    }
  }
  if (a1 == 1 && b1 == 0) {
    if (crr[i].second == -1)
      crr[i + 1].second = -1;
    else
      crr[i + 1].second = 1 ^ crr[i].second;
  } else {
    if (a1 == 1 && b1 == 1) {
      if (crr[i].second == 0)
        return false;
      else
        crr[i].second = crr[i + 1].second = 1;
    } else {
      if (a1 == 0 && b1 == 1)
        return false;
      else {
        if (crr[i].second == 1)
          return false;
        else
          crr[i].second = crr[i + 1].second = 0;
      }
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  int arr[100010], brr[100010];
  pair<int, int> crr[100010];
  for (int i = 1; i < n; ++i) {
    cin >> arr[i];
    crr[i].first = crr[i].second = -1;
  }
  for (int i = 1; i < n; ++i) cin >> brr[i];
  crr[n].first = crr[n].second = -1;
  bool poss = true;
  for (int i = 1; i < n; ++i) {
    if (!fun(arr, brr, crr, i)) {
      poss = false;
      break;
    }
  }
  if (poss) {
    crr[n + 1].first = crr[n + 1].second = 0;
    cout << "YES" << endl;
    for (int i = n; i >= 1; --i) {
      if (crr[i].first == -1) {
        crr[i].first = 1 ^ crr[i + 1].first;
      }
      if (crr[i].second == -1) {
        crr[i].second = 1 ^ crr[i + 1].second;
      }
    }
    for (int i = 1; i <= n; ++i)
      cout << crr[i].second * 2 + crr[i].first << " ";
  } else
    cout << "NO" << endl;
}
