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()))

pos_and = [
    {(0, 1), (0, 2), (0, 3), (0, 0), (1, 2)},
    {(1, 3), (1, 1)},
    {(2, 3), (2, 2)},
    {(3, 3)}
]
pos_or = [
    {(0, 0)},
    {(0, 1), (1, 1)},
    {(0, 2), (2, 2)},
    {(0, 3), (1, 3), (2, 3), (3, 3), (1, 2)}
]

# for i in range(4):
#     for j in range(4):
#         print(i, j, [x for x in pos_and[i].intersection(pos_or[j])])

res1 = []
flag1 = False
for i in range(n - 1):
    pairs = [x for x in pos_and[b[i]].intersection(pos_or[a[i]])]
    pairs.sort()
    if len(pairs) == 0:
        flag1 = True
        break
    for pair in pairs:
        rev_pair = (pair[1], pair[0])
        if len(res1) == 0:
            res1.append(pair[0])
            res1.append(pair[1])
            break
        if res1[-1] == pair[0]:
            res1.append(pair[1])
            break
        elif res1[-1] == rev_pair[0]:
            res1.append(rev_pair[1])
            break
    if len(res1) != i + 2:
        flag1 = True
        break
res2 = []
flag2 = False
for i in range(n - 1):
    pairs = [x for x in pos_and[b[i]].intersection(pos_or[a[i]])]
    pairs.sort()
    if len(pairs) == 0:
        flag2 = True
        break
    for pair in pairs:
        rev_pair = (pair[1], pair[0])
        if len(res2) == 0:
            res2.append(rev_pair[0])
            res2.append(rev_pair[1])
            break
        if res2[-1] == pair[0]:
            res2.append(pair[1])
            break
        elif res2[-1] == rev_pair[0]:
            res2.append(rev_pair[1])
            break
    if len(res2) != i + 2:
        flag2 = True
        break
if flag1 and flag2 and a[0] == 3 and b[0] == 0:
    res1 = []
    flag1 = False
    for i in range(n - 1):
        pairs = [x for x in pos_and[b[i]].intersection(pos_or[a[i]])]
        pairs.sort()
        if len(pairs) == 0:
            flag1 = True
            break
        for pair in pairs:
            rev_pair = (pair[1], pair[0])
            if len(res1) == 0:
                res1.append(pairs[1][0])
                res1.append(pairs[1][1])
                break
            if res1[-1] == pair[0]:
                res1.append(pair[1])
                break
            elif res1[-1] == rev_pair[0]:
                res1.append(rev_pair[1])
                break
        if len(res1) != i + 2:
            flag1 = True
            break
    res2 = []
    flag2 = False
    for i in range(n - 1):
        pairs = [x for x in pos_and[b[i]].intersection(pos_or[a[i]])]
        pairs.sort()
        if len(pairs) == 0:
            flag2 = True
            break
        for pair in pairs:
            rev_pair = (pair[1], pair[0])
            if len(res2) == 0:
                res2.append(pairs[1][1])
                res2.append(pairs[1][0])
                break
            if res2[-1] == pair[0]:
                res2.append(pair[1])
                break
            elif res2[-1] == rev_pair[0]:
                res2.append(rev_pair[1])
                break
        if len(res2) != i + 2:
            flag2 = True
            break
if not flag2:
    print("YES")
    print(*res2)
elif not flag1:
    print("YES")
    print(*res1)
else:
    print("NO")
