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
from collections import defaultdict
def solve(N,A,B):
    d1=defaultdict(int)
    d2=defaultdict(int)
    for i in range(N):
        d1[A[i]]+=1
        d2[B[i]]+=1
    for i in d1:
        if d1[i]+d2[i]>N:
            return "No",[]
    else:
        return "Yes",B[::-1]
    
if __name__ == "__main__":
    N=int(input())
    A=[int(i) for i in input().split()][:N]
    B=[int(i) for i in input().split()][:N]
    a,b=solve(N,A,B)
    print(a)
    if b!=[]:
        #print(a)
        k,z=0,N-1
        for i in range(N):
            if A[i]==b[i] and b[i]!=b[k]:
                b[i],b[k]=b[k],b[i]
                k+=1
            elif A[i]==b[i] and b[i]==b[k]:
                b[i],b[z]=b[z],b[i]
                z-=1
        z,k=0,N-1
        for i in range(N):
            if A[i]==b[i] and b[i]!=b[k]:
                b[i],b[k]=b[k],b[i]
                k-=1
            elif A[i]==b[i] and b[i]==b[k]:
                b[i],b[z]=b[z],b[i]
                z+=1
        print(*b)
