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
#!/usr/bin/env pypy3

def ans(N, A, B):
    B = B[::-1]

    match_vals = set()

    for i in range(N):
        if A[i] == B[i]:
            match_vals.add(A[i])

    if len(match_vals) == 0:
        print("Yes")
        print(*B)
        return

    assert(len(match_vals) == 1)
    [m] = list(match_vals)

    ok = []
    bad = []
    for i in range(N):
        if B[i] != m and A[i] != m:
            ok += [i]
        if A[i] == B[i]:
            bad += [i]

    if len(bad) > len(ok):
        print('No')
        return

    for i in range(len(bad)):
        B[bad[i]], B[ok[i]] = B[ok[i]], B[bad[i]]

    print('Yes')
    print(*B)

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

ans(N, A, B)