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
from collections import Counter, deque

(N,) = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
B = [int(x) for x in input().split()]

freqA = Counter(A)
freqB = Counter(B)
for k in freqA.keys() & freqB.keys():
    if freqA[k] + freqB[k] > N:
        print("No")
        exit()

B = deque(B)
while True:
    rotate = 0
    for i in range(N):
        while A[(i + rotate) % N] == B[i]:
            rotate += 1
    bad = False
    B.rotate(rotate)
    for a, b in zip(A, B):
        if a == b:
            bad = True
    if not bad:
        break

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

