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;
long long power(long long a, long long b) {
  long long res = 1ll;
  while (b > 0) {
    if (b % 2 != 0) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return res;
}
long long ncr(long long n, long long k) {
  if (k > n - k) k = n - k;
  long long pro = 1;
  for (long long i = 0; i < k; i++) {
    pro = (pro * (n - i)) % 1000000007;
    pro /= (i + 1);
  }
  return (pro % 1000000007);
}
vector<long long> prime;
void seive() {
  vector<bool> isprime(2000001, true);
  for (int i = 2; i * i <= 2000000; i++) {
    if (isprime[i]) {
      for (int j = i * i; j <= 2000000; j += i) {
        isprime[j] = false;
      }
    }
  }
  for (int i = 2; i <= 1000000; i++) {
    if (isprime[i]) prime.push_back(i);
  }
}
void swap(long long *a, long long *b) {
  long long temp = *a;
  *a = *b;
  *b = temp;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n;
  vector<long long> 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];
  }
  int pos1, pos2;
  for (int i = 0; i <= 3; i++) {
    for (int j = 0; j <= 3; j++) {
      int flag1 = 0;
      if ((i & j) == b[0] && (i | j) == a[0]) {
        pos1 = i;
        pos2 = j;
        long long pos22 = pos2;
        for (int k = 1; k < n - 1; k++) {
          int flag3 = 0;
          for (int l = 0; l <= 3; l++) {
            if ((pos22 | l) == a[k] && (pos22 & l) == b[k]) {
              pos22 = l;
              flag3 = 1;
              break;
            }
          }
          if (flag3 == 0) {
            flag1 = 1;
            break;
          }
        }
        if (flag1 == 0) {
          cout << "YES" << endl;
          cout << pos1 << " " << pos2 << " ";
          for (int i = 1; i < n - 1; i++) {
            for (int j = 0; j <= 3; j++) {
              if ((pos2 | j) == a[i] && (pos2 & j) == b[i]) {
                pos2 = j;
                cout << pos2 << " ";
                break;
              }
            }
          }
          return 0;
        }
      }
    }
  }
  cout << "NO";
  return 0;
}
