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.
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
ended = False
def toBIN(x):
    if x == 3:
        return'11'
    elif x == 2:
        return '10'
    elif x == 1:
        return '01'
    else:
        return '00'
def toDEC(x):
    if x == '00':
        return 0
    elif x == '01':
        return 1
    elif x == '10':
        return 2
    else:
        return 3
def _or(t_now1, t_now2):
    x = ''
    if t_now1[0] == '0' and t_now2[0] == '0':
        x = '0'
    else:
        x = '1'
    if t_now1[1] == '0' and t_now2[1] == '0':
        x += '0'
    else:
        x += '1'
    return x
def _and(t_now1, t_now2):
    x = ''
    if t_now1[0] == '1' and t_now2[0] == '1':
        x = '1'
    else:
        x = '0'
    if t_now1[1] == '1' and t_now2[1] == '1':
        x += '1'
    else:
        x += '0'
    return x
def newT(mas):
    if len(mas) == n:
        print('YES')
        ended = True
        for u in mas:
            print(toDEC(u), end=' ')
        exit()
    else:
        t = mas
        t_now1 = t[len(t)-1]
        for j in range(4):
            mas = t
            t_now2 = toBIN(j)
            if len(t)-1 <= len(a)-1:
                if (toBIN(a[len(t)-1]) == _or(t_now1, t_now2)) and (toBIN(b[len(t)-1] == _and(t_now1, t_now2))):
                    mas.append(t_now2)
                    newT(mas)
for i in range(4):
    t_now1 = toBIN(i)
    for j in range(4):
        t_now2 = toBIN(j)
        if (toBIN(a[0]) == _or(t_now1, t_now2)) and (toBIN(b[0]) == _and(t_now1, t_now2)):
            newT([t_now1, t_now2])
if not ended:
    print('NO')