Given are two sequences A and B, both of length N. A and B are each sorted in the ascending order. Check if it is possible to reorder the terms of B so that for each i (1 \leq i \leq N) A_i \neq B_i holds, and if it is possible, output any of the reorderings that achieve it.

Constraints

* 1\leq N \leq 2 \times 10^5
* 1\leq A_i,B_i \leq N
* A and B are each sorted in the ascending order.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 \cdots A_N
B_1 B_2 \cdots B_N


Output

If there exist no reorderings that satisfy the condition, print `No`.

If there exists a reordering that satisfies the condition, print `Yes` on the first line. After that, print a reordering of B on the second line, separating terms with a whitespace.

If there are multiple reorderings that satisfy the condition, you can print any of them.

Examples

Input

6
1 1 1 2 2 3
1 1 1 2 2 3


Output

Yes
2 2 3 1 1 1


Input

3
1 1 2
1 1 3


Output

No


Input

4
1 1 2 3
1 2 3 3


Output

Yes
3 3 1 2
n=int(input())
y=lambda:[*map(int,input().split())]
a,b=y(),y()[::-1]
c=0
for i in range(n):
  if a[i]==b[i]:c=a[i]
if c<1:print("Yes");print(' '.join(map(str,b)));quit()
al,bl,ah,bh=[-1]*4
ea=eb=0
for i in range(n):
  if al<0 and a[i]==c:al=i
  if bl<0 and b[i]==c:bl=i
  if al<0:ea+=1
  if bl<0:eb+=1
  if ah<0 and al>=0 and a[i]!=c:ah=i
  if bh<0 and bl>=0 and b[i]!=c:bh=i
if ah<0:ah=n
if bh<0:bh=n
t=ah+bh-al-bl
if ah+bh-al-bl>n:print("No");quit()
o1=min(bh-bl,ea,ea-eb+ah-al)
print("Yes")
print(' '.join(map(str,[c]*o1+b[:eb]+b[eb+bh-bl:]+[c]*(bh-bl-o1))))