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 sys
readline = sys.stdin.readline

N = int(readline())
A = list(map(int,readline().split()))
B = list(map(int,readline().split()))

from collections import Counter

cA = Counter(A)
cB = Counter(B)

# 両チームに入っている文字の合計数が多い順に処理していく
# heapqで、(両方に入っている個数, Bにいる個数, その数)
# 使うごとに、1つ目、2つ目の要素を-1してもう一度heappush

import heapq as hq
q = []
hq.heapify(q)
for key,value in cB.items():
  if key in cA:
    hq.heappush(q, (-(value + cA[key]), value, key))
    if value + cA[key] > N:
      print("No")
      exit(0)
  else:
    hq.heappush(q, (-value, value, key))
print("Yes")
a_ind = 0
ans = []

while a_ind < N:
  keep = []
  while True:
    allcnt, bcnt, key = hq.heappop(q)
    allcnt *= (-1)
    if A[a_ind] == key:
      keep.append((allcnt, bcnt, key))
    else:
      ans.append(key)
      if bcnt - 1 > 0:
        hq.heappush(q, (-(allcnt - 1), bcnt - 1, key))
      break
  for k in keep:
    hq.heappush(q, k)
  a_ind += 1
    
print(*ans)