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 Counter
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
AC = Counter(A)
BC = Counter(B)
for i in BC.keys():
    if i in AC and BC[i] + AC[i] > N:
        print("No")
        exit()
AL,AR = {},{} 
BL,BR = {},{}
for i in range(N):
    if not A[i] in AL:
        AL[A[i]] = i
    AR[A[i]] = i

for i in range(N):
    if not B[i] in BL:
        BL[B[i]] = i
    BR[B[i]] = i
can = [1]*(N)

for i in BL.keys():
    if not i in AL:
        continue
    s = (N+AL[i]-BR[i])%N
    e = (N+AR[i]-BL[i])%N
    if e<s:
        e += N
    # print(i,s,e)
    for j in range(s,e+1):
        can[j%N] = 0

if 1 in can:
    print("Yes")
    ind = can.index(1)
    print(* (B+B)[N-ind:2*N-ind])
else:
    print("No")
