Problem: 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

Solution (in PYTHON3):

Solution:
N = int(input())
from collections import Counter
A = list(map(int, input().split()))
B = list(map(int, input().split()))
ca = Counter(A)
cb = Counter(B)
mod = 10**6
cnts = [set() for _ in range(N+1)]
for i in range(1,N+1):
  if ca[i]+cb[i]>N:
    print('No')
    exit()
  if ca[i]+cb[i]>0:
    cnts[ca[i]+cb[i]].add(i)

def get_any(counts,r):
  for k in counts.keys():
    if counts[k]>0 and k!=r:
      return k

def update_cnt(cnt_,r):
  cnt_[r] -= 1
  if cnt_[r]==0:
    del cnt_[r]

lis = [[] for _ in range(N+1)]
for i in range(N,0,-1):
  if len(cnts[i])==2:
    p,q = cnts[i]
    for j in range(ca[p]):
      lis[p].append(q)
    for j in range(ca[q]):
      lis[q].append(p)
    break
  elif len(cnts[i])==1:
    p, = cnts[i]
    for j in range(ca[p]):
      q = get_any(cb,p)
      update_cnt(cb,q)
      lis[p].append(q)
    for j in range(cb[p]):
      q = get_any(ca,p)
      update_cnt(ca,q)
      lis[q].append(p)
    break
  else:
    p = get_any(ca,-1)
    q = get_any(cb,p)
    lis[p].append(q)
    cnts[ca[p]+cb[p]].remove(p)
    cnts[ca[q]+cb[q]].remove(q)
    update_cnt(ca,p)
    update_cnt(cb,q)
    cnts[ca[p]+cb[p]].add(p)
    cnts[ca[q]+cb[q]].add(q)
ans = []
for i in range(1,N+1):
  for b in lis[i]:
    ans.append(b)
print('Yes')
print(*ans)