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()))

CA = [0] * (N + 1)
CB = [0] * (N + 1)

for a in A:
  CA[a] += 1
for b in B:
  CB[b] += 1
X = 0
x = 0
y = 0
ans = [[0, 0] for i in range(N)]

for i in range(1, N + 1):
  if CA[i] + CB[i] > N:
    print('No')
    exit()
  if X + CA[i] + CB[i] < N:
    for j in range(X, X + CA[i]):
      ans[j][0] = i
    for j in range(X + CA[i], X + CA[i] + CB[i]):
      ans[j][1] = i
    X += CA[i] + CB[i]
    x += CA[i]
    y += CB[i]
    CA[i] = 0
    CB[i] = 0
  else:
    k = min(N, N + x - CB[i], X + CA[i])
    for j in range(X, k):
      ans[j][0] = i
    for j in range(k, N):
      ans[j][1] = i
    CA[i] -= k - X
    CB[i] -= N - k
    break
x = 1
y = 1
for j in range(N):
  while x < N + 1 and CA[x] == 0:
    x += 1
  if ans[j][0] == 0:
    ans[j][0] = x
    CA[x] -= 1
for j in range(N):
  while y < N + 1 and CB[y] == 0:
    y += 1
  if ans[j][1] == 0:
    ans[j][1] = y
    CB[y] -= 1
ans.sort()
print('Yes')
print(*(ans[i][1] for i in range(N)))
  