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
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
c = [0 for _ in range(n+1)]
for ai in a:
  c[ai] += 1
for bi in b:
  c[bi] += 1
if max(c) > n:
  print("No")
  exit()
print("Yes")
b.reverse()
flag = True
si = -1
ti = -1
for i in range(n):
  if flag:
    if a[i] == b[i]:
      si = i
      flag = False
  else:
    if a[i] != b[i]:
      ti = i-1
      break
if si == -1:
  for bi in b:
    print(bi,end=" ")
  exit()
if ti == -1:
  ti = n-1
s = a[si]
ni = [0 for _ in range(ti-si+1)]
xi = 0
for i in range(n):
  if xi >= ti-si+1:
    break
  if a[i] != s and b[i] != s:
    ni[xi] = b[i]
    b[i] = s
    xi += 1
nb = b[0:si] + ni + b[ti+1:]
for nbi in nb:
  print(nbi,end=" ")