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
import heapq
from collections import defaultdict
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))

d = defaultdict(list)
for i in range(N):
    a = A[i]
    b = B[i]
    if a==b:
        d[a].append(i)

h = []
for a in d:
    heapq.heappush(h,(-len(d[a]),a))

while len(h)>=2:
    x,a = heapq.heappop(h)
    y,b = heapq.heappop(h)
    i = d[a].pop()
    j = d[b].pop()
    B[i],B[j] = B[j],B[i]
    x += 1
    y += 1
    if x<0:
        heapq.heappush(h,(x,a))
    if y<0:
        heapq.heappush(h,(y,b))

if h:
    x,a = heapq.heappop(h)
    d[a] = d[a][::-1]
    for i in range(N):
        if len(d[a])==0:
            break
        if A[i] != a and B[i] != a:
            j = d[a].pop()
            B[i],B[j] = B[j],B[i]

ans = "Yes"
for i in range(N):
    if A[i]==B[i]:
        ans = "No"

print(ans)
if ans=="Yes":
    print(*B)
