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

A1 = collections.Counter(A)
B1 = collections.Counter(B)
A2 = A1.most_common()
B2 = B1.most_common()

if A2[0][0] == B2[0][0]:
  if min(A2[0][1], B2[0][1]) > int(N / 2):
    print("No")
    quit()

B.reverse()    
s = 0
t = N - 1
for i in range(N):
  if A[i] == B[i]:
    if (A[s] != B[i]) and (A[i] != B[s]):
      x = B[s]
      B[s] = B[i]
      B[i] = x
      s += 1
    elif (A[t] != B[i]) and (A[i] != B[t]) :
      x = B[t]
      B[t] = B[i]
      B[i] = x
      t -= 1
    else:
      print("No")
      quit()
  
#print(B)  
print("Yes")  
for i in range(N):
  print(B[i], end = " ")
print("")

  
    
