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 bisect
import collections
import copy
import functools
import heapq
import math
import sys
from collections import deque
from collections import defaultdict
from fractions import Fraction
input = sys.stdin.readline
sys.setrecursionlimit(10**9)
INF = float("inf")
MOD = 10**9+7

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

B.reverse()

for i in range(N):
    if A[i] == B[i]:
        k = A[i]
        break
else:
    print("Yes")
    print(*B)
    exit()

possible = []
impossible = []
for i in range(N):
    if A[i] != k and B[i] != k:
        possible.append(i)
    elif A[i] == k and B[i] == k:
        impossible.append(i)

for i in impossible:
    if not possible:
        print("No")
        exit()
    j = possible.pop()
    B[i],B[j] = B[j],B[i]

print("Yes")
print(*B)