Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

6
1 1 1 2 2 3
1 1 1 2 2 3

Sample Output 1

Yes
2 2 3 1 1 1

Sample Input 2

3
1 1 2
1 1 3

Sample Output 2

No

Sample Input 3

4
1 1 2 3
1 2 3 3

Sample Output 3

Yes
3 3 1 2

Solution in OCaml (4.10.0) (judge verdict: Accepted)

Scanf.scanf "%d" (fun n ->
    let a = Array.make (n + 1) 0 in
    let b = Array.make (n + 1) 0 in
    let bb = Array.make n 0 in
    let module S = Set.Make (struct type t = int * int let compare = compare end) in
    for i = 1 to n do Scanf.scanf " %d" (fun j -> a.(j) <- a.(j) + 1) done;
    for i = 1 to n do Scanf.scanf " %d" (fun j -> b.(j) <- b.(j) + 1; bb.(i - 1) <- j) done;

    let rec check i =
        if i > n then true else if a.(i) + b.(i) > n then false else check (i + 1)
    in
    if check 1 then (
        let c = Array.make (n + 1) 0 in
        let d = Array.make (n + 1) 0 in
        for i = 1 to n do
            c.(i) <- c.(i - 1) + a.(i);
            d.(i) <- d.(i - 1) + b.(i);
        done;
        let rec loop i mx =
            if i > n then mx else loop (i + 1) (max mx (c.(i) - d.(i - 1)))
        in
        let x = loop 1 0 in
        print_endline "Yes";
        for i = 0 to n - 1 do
            Printf.printf "%d " bb.((i - x + n) mod n)
        done;
        print_newline ()
    ) else print_endline "No"
)