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
n = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = Counter(A)
D = Counter(B)
for k, v in C.items():
    if v + D[k] > n:
        print("No")
        exit()
print("Yes")
seen = set()
L = []
E = dict()
for i, a in enumerate(A):
    if a not in seen:
        seen.add(a)
        if i != 0:
            E[A[i-1]] = i-1
E[A[-1]] = -1
seen = [False]*n
K = [-1]*n
M = []
for k, v in D.items():
    if k in C:
        M.append((C[k]+v, k))
M.sort(reverse=True)
for p, q in M:
    if q in D:
        s = D[q]
        i = E[q]
        for j in range(s):
            if seen[(i+j+1)%n]:
                break
            seen[(i+j+1)%n] = True
            K[(i+j+1)%n] = q
            D[q] -= 1
        else:
            D.pop(q)
            
            
for i in range(n-1, -1, -1):
    if K[i] == -1:
        for k in D.keys():
            if A[i] != k:
                K[i] = k
                break
        D[k] -= 1
        if D[k] == 0:
            D.pop(k)
print(*K)