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 sys
input = lambda: sys.stdin.readline().rstrip()
N = int(input())
A = [int(a)-1 for a in input().split()]
B = [int(a)-1 for a in input().split()]

LA = [0] * N + [N]
LB = [0] * N + [N]
j = 0
for i in range(N):
    LA[i] = j
    while j < N and A[j] == i:
        j += 1
j = 0
for i in range(N):
    LB[i] = j
    while j < N and B[j] == i:
        j += 1

NG = [0] * (N + 1)
for l1, r1, l2, r2 in zip(LA[:-1], LA[1:], LB[:-1], LB[1:]):
    ngl = l1-r2+1
    ngr = r1-l2
    if ngl < ngr < 0:
        NG[ngl+N] += 1
        NG[ngr+N] -= 1
    elif ngl < 0 <= ngr:
        NG[ngl+N] += 1
        NG[N] -= 1
        NG[0] += 1
        NG[ngr] -= 1
    elif 0 <= ngl < ngr:
        NG[ngl] += 1
        NG[ngr] -= 1
for i in range(N):
    NG[i+1] += NG[i]

for i in range(N):
    if NG[i]: continue
    print("Yes")
    print(*[a + 1 for a in (B[-i:] + B[:-i] if i else B)])
    break
else:
    print("No")