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
from collections import Counter
from collections import defaultdict

sys.setrecursionlimit(10 ** 7)

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

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

CA = Counter(A)
CB = Counter(B)

right_idx = defaultdict(lambda: -1)
left_idx = defaultdict(lambda: -1)

for i in range(N):
    a = A[i]
    right_idx[a] = i

for i in range(N)[::-1]:
    b = B[i]
    left_idx[b] = i

# check
for X in [A, B]:
    for x in X:
        if CA[x] + CB[x] > N:
            print("No")
            exit()


shift = 0
for x in range(1, N + 1):
    if right_idx[x] == -1 or left_idx[x] == -1:
        continue

    shift = max(right_idx[x] - left_idx[x] + 1, shift)


ANS = B[-shift:] + B[:-shift]

print("Yes")
print(" ".join(map(str, ANS)))

