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 11.0.6) (judge verdict: Runtime Error)

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Comparator;
import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.Vector;

import javax.swing.JTabbedPane;
public class Main{
	static class FastScanner{//用于快速读入大量数据
		BufferedReader br;
		StringTokenizer st;
		public FastScanner(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in),16384);
			eat("");
		}
		public void eat(String s) {
			st = new StringTokenizer(s);
		}
		public String nextLine() {
			try {
				return br.readLine();
			} catch (IOException e) {
				return null;
			}
		}
		public boolean hasNext() {
			while(!st.hasMoreTokens()) {
				String s = nextLine();
				if(s==null) return false;
				eat(s);
			}
			return true;
		}
		public String next() {
			hasNext();
			return st.nextToken();
		}
	    
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
	static FastScanner cin = new FastScanner(System.in);//快读
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));//快速输出
    static int mod=1000000007;

   


   static  String str;
    public static void main(String[] args) {
       int n;
       n=cin.nextInt();
       str=cin.next();
       int len=str.length();
       int ans=0;
       Vector<Integer>vec=new Vector<Integer>();
       for(int i=len-1;i>=0;i--) {
    	   if(str.charAt(i)=='R') {
    		   vec.add(i);
    	   }
       } 
       int cc=0;
       for(int i=0;i<len;i++) {
    	   if(str.charAt(i)=='W') {
    		   if(vec.elementAt(cc)>i) {
    			   cc++;
    			   ans++;
    		   }
    	   }
       }
       out.println(ans);
       out.flush();
       return ;
	}
 
}