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.
#!/usr/bin/env python3

bits = {0: (0,0), 1: (0,1), 2: (1,0), 3: (1,1)}
nums = {(0,0): 0, (0,1): 1, (1,0): 2, (1,1): 3}
var = {(0,0): [(0,0)], (0,1): [], (1,0): [(1,0), (0,1)], (1,1): [(1,1)]}

def intersection(lst1, lst2): 
	lst3 = [value for value in lst1 if value in lst2] 
	return lst3

def getarr(a, b, n):
	arr = [[]] * (n-1)
	for i in range(0,n-1):
		arr[i] = var[(a[i], b[i])]

	return arr

def gett(arr, n):
	t = [[]] * n
	for i in range(1,n-1):
		inter = intersection([x[1] for x in arr[i-1]], [x[0] for x in arr[i]])
		if inter == []:
			print('NO')
			exit()
		t[i] = inter[0]

	for x in arr[0]:
		if t[1] == x[1]:
			t[0] = x[0]

	for x in arr[n-2]:
		if t[n-2] == x[0]:
			t[n-1] = x[1]

	return t


n = int(input())
a = [bits[int(x)] for x in input().split(" ")]
b = [bits[int(x)] for x in input().split(" ")]

fa = [x[0] for x in a]
sa = [x[1] for x in a]
fb = [x[0] for x in b]
sb = [x[1] for x in b]
if n != 2:
	farr = getarr(fa, fb, n)
	sarr = getarr(sa, sb, n)

	if ([] in farr) or ([] in sarr):
		print('NO')
		exit()

	ft = gett(farr, n)
	st = gett(sarr, n)

	t = [(ft[i],st[i]) for i in range(0, n)]
	t = [str(nums[x]) for x in t]

	print('YES')
	print(' '.join(t))
else:
	farr = getarr(fa, fb, n)
	sarr = getarr(sa, sb, n)

	if ([] in farr) or ([] in sarr):
		print('NO')
		exit()

	ft = [farr[0][0][0], farr[0][0][1]]
	st = [sarr[0][0][0], sarr[0][0][1]]
	t = [(ft[i],st[i]) for i in range(0, n)]
	t = [str(nums[x]) for x in t]

	print('YES')
	print(' '.join(t))