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 Python (3.8.2) (judge verdict: Runtime Error)

n = int(input())
input_str = input()
first_r = input_str.find("R")
r_num = input_str[first_r:].count("R")
if first_r == -1:
    print(0)
    exit()
if input_str.find("R").count("W"):
    print(0)
    exit()
if first_r > r_num:
    print(r_num)
    exit()
count = 0
for i in range(n):
    if input_str[-(i+1)] == "R":
        count += 1
    if count == first_r:
        index_lr = n-1-i
        break
first_w = input_str[first_r:].find("W")
Lr = input_str[first_r:index_lr].rfind("R")
count_r = input_str[first_w+first_r:].count("R")
count_w = input_str[first_r:first_r+Lr].count("W")
if count_r-first_r > count_w:
    print(count_w+first_r)
else:
    print(count_r)