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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

N = int(input())
c = input()

count = 0
rui_red = [0 for _ in range(N)]
rui_white = [0 for _ in range(N)]
now_red = 0
now_white = 0
for i in range(N):
    if c[i] == "W":
        now_white += 1
        rui_white[i] = now_white
    if c[-1-i] == "R":
        now_red += 1
        rui_red[-1-i] = now_red

left = 0
right = N-1
for i in range(N):
    pre_left = left
    pre_right = right
    
    while rui_white[left] < i:
        left += 1
        if  left >= N:
             break
    while rui_red[right] < i:
        right -= 1
        if right < 0:
            break
    
    if left>=right:
        break
answer = i-1
count = 0
for i in range(pre_left+1, pre_right):
    if c[i] == "W":
        count+=1
    else:
        count-=1

print(answer)