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
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
cnt = [0 for i in range(n+1)]

def lb(x):
    l, r = 0, n
    while l < r:
        m = (l+r)//2
        if a[m] < x:
            l = m + 1
        else:
            r = m
    return l

def ub(x):
    l, r = 0, n
    while l < r:
        m = (l+r)//2
        if a[m] <= x:
            l = m + 1
        else:
            r = m
    return l

for i in range(n):
    l = lb(b[i])
    r = ub(b[i])-1
    if l > r:
        continue
    if l < i and i <= r:
        cnt[l+n-i]+=1
        cnt[0]+=1
        cnt[r-i+1]-=1
    elif i <= l:
        cnt[l-i]+=1
        cnt[r-i+1]-=1
    elif i > r:
        cnt[l+n-i]+=1
        cnt[r+n+1-i]-=1

res = 0
ans = -1
for i in range(n):
    res += cnt[i]
    if(res == 0):
        ans = i
if ans == -1:
    print("No")
else:
    print("Yes")
    for i in range(n):
        print(b[(i-ans+n)%n],end=' ')