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())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
coa=[0]*(n+1)
cob=[0]*(n+1)
for i in a:
  coa[i]+=1
for i in b:
  cob[i]+=1
for i in range(n+1):
  if coa[i]+cob[i]>n:print("No");exit()
b.sort(reverse=True)
ans = b[:]
changed=0
for i in range(n):
  if a[i]==b[i]:
    ans[i],ans[changed]=ans[changed],ans[i]
    changed+=1
chenged_b=-1
for i in range(n):
  if a[i]==ans[i]:
    ans[i],ans[chenged_b]=ans[chenged_b],ans[i]
    chenged_b-=1
print("Yes")
print(*ans)