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()))
B.sort(reverse=True)
Arev=sorted(A, reverse=True)
Brev=sorted(B)
atai=0
s=0
for i in range(len(A)):
    if A[i]==B[i]:
        atai=A[i]
        s+=1
C=[]
for i in range(len(A)):
    C.append(A[i]-B[i])
if s>0:
    hajime=C.index(0)
if s==0:
    print("Yes")
    print(*B)
else:
    Acount=A.count(atai)
    Bcount=B.count(atai)
    if N-Acount<Bcount:
        print("No")
    else:
        afirst=A.index(atai)
        alast=N-1-Arev.index(atai)
        bfirst=B.index(atai)
        blast=N-1-Brev.index(atai)
        t=min(afirst,bfirst)
        if t>=s:
            for i in range(0,s):
                B[hajime+i]=B[i]
                B[i]=atai
        else:
            for i in range(0,t):
                B[hajime+i]=B[i]
                B[i]=atai
            ss=s-t
            for i in range(0,ss):
                B[hajime+t+i]=B[-i-1]
                B[-i-1]=atai
        print("Yes")
        print(*B)