Problem:
Score : 400 points

Problem Statement

An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; R stands for red and W stands for white.

You can do the following two kinds of operations any number of times in any order:

Choose two stones (not necessarily adjacent) and swap them.

Choose one stone and change its color (from red to white and vice versa).

According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?

Constraints

2 \leq N \leq 200000

c_i is R or W.

Input

Input is given from Standard Input in the following format:

N
c_{1}c_{2}...c_{N}

Output

Print an integer representing the minimum number of operations needed.

Sample Input 1

4
WWRR

Sample Output 1

2

For example, the two operations below will achieve the objective.

Swap the 1-st and 3-rd stones from the left, resulting in RWWR.

Change the color of the 4-th stone from the left, resulting in RWWW.

Sample Input 2

2
RR

Sample Output 2

0

It can be the case that no operation is needed.

Sample Input 3

8
WRWWRWRR

Sample Output 3

3

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

let () = Scanf.scanf "%d\n" @@ fun n ->
  let cs = Array.init n @@ fun _ -> Scanf.scanf "%c" Fun.id in
  let reds = Array.make (n + 1) 0 in (* th以上の赤 *)
  let whites = Array.make (n + 1) 0 in (* th未満の白 *)
  for i = 0 to n - 1 do
    match cs.(i) with
    | 'R' -> reds.(i) <- reds.(i) + 1
    | 'W' -> whites.(i + 1) <- whites.(i + 1) + 1
  done;
  for i = n - 1 downto 0 do
    reds.(i) <- reds.(i) + reds.(i + 1)
  done;
  for i = 0 to n - 1 do
    whites.(i + 1) <- whites.(i) + whites.(i + 1)
  done;
  Printf.printf "%d\n" @@
  Array.fold_left min max_int @@
  Array.init (n + 1) @@ fun i -> max reds.(i) whites.(i)