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 Java (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = Integer.parseInt(sc.next());
		int count =0;
		int front = 0;
		int last = N-1;
		String c = sc.next();
		StringBuilder cc = new StringBuilder(c);
		while(true) {
			label1: for(int i=front; i<last; i++) {
				if(cc.substring(i,i+1).equals("W") && cc.substring(i+1,i+2).equals("R")) {
					for(int j=last; j>i;j--) {
						if(cc.substring(j,j+1).equals("R")) {
							cc.setCharAt(i, 'R');
							cc.setCharAt(j, 'W');
//							front = i;
//							last = j;
							i = front-1;
							j = j-1;
							count++;
							break ;
						}
					}
				}
			}
					System.out.println(count);
					return;
		}

	}
}