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
def main():
    from collections import Counter
    N = int(input())
    A = list(map(int,input().split()))
    B = list(map(int,input().split()))
    a_count = Counter(A)
    b_count = Counter(B)
    for k,v in (a_count + b_count).items():
        if N < v:
            print('No')
            return
    print('Yes')
    C = [0]*(N+1)
    idx = 0
    for a in A:
        while idx < a:
            idx += 1
            C[idx] = C[idx-1]
        C[idx] += 1
    while idx < N:
        idx += 1
        C[idx] = C[idx-1]

    D = [0]*(N+1)
    idx = 0
    for b in B:
        while idx < b:
            idx += 1
            D[idx] = D[idx-1]
        D[idx] += 1
    while idx < N:
        idx += 1
        D[idx] = D[idx-1]
    
    x = 0
    for i in range(1,N+1):
        x = max(x, C[i] - D[i-1])
        # print(x)
    print(*(B[N-x:]+B[:N-x]))
    # print(C)
    # print(D)

main()
