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
from heapq import heappush, heappop, heapreplace


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

AC = Counter(A)
BC = Counter(B)
S = AC + BC

q = []
for k, v in S.items():
    heappush(q, (-v, k))

ans = []
while q:
    v, k = q[0]
    v = -v
    if v == 0:
        heappop(q)
    elif v != S[k]:
        heapreplace(q, (-S[k], k))
    else:
        heapreplace(q, (-(v - 1), k))
        if v > N:
            print('No')
            exit()
        else:
            if AC[k] > 0:
                a = k
                for l in BC.keys():
                    if l != k:
                        b = l
                        break
            else:
                b = k
                for l in AC.keys():
                    if l != k:
                        a = l
                        break
            ans.append((a, b))
            S[a] -= 1
            S[b] -= 1
            AC[a] -= 1
            if not AC[a]:
                AC.pop(a)
            BC[b] -= 1
            if not BC[b]:
                BC.pop(b)

print('Yes')
ans.sort()
X = [b for a, b in ans]
print(*X)
