Problem:
Score : 300 points

Problem Statement

There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.

Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.

Ultimately, how many slimes will be there?

Constraints

1 \leq N \leq 10^5

|S| = N

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the final number of slimes.

Sample Input 1

10
aabbbbaaca

Sample Output 1

5

Ultimately, these slimes will fuse into abaca.

Sample Input 2

5
aaaaa

Sample Output 2

1

All the slimes will fuse into one.

Sample Input 3

20
xxzaffeeeeddfkkkkllq

Sample Output 3

10

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;

class Main {
    static Scanner scanner = new Scanner();
    static long mod = 1000000007;
    static int inf = 0x7ffffff;

    public static void main(String[]$) throws Exception {
        int n = scanner.nextInt();
        char[] c = scanner.next().toCharArray();
        int a = n;
        for (int i = 1; i < n; i++) {
            if (c[i - 1] == c[i]) {
                a--;
            }
        }
        System.out.println(a);
    }

    static class System {
        static PrintWriter out = new PrintWriter(java.lang.System.out);

        static {
            Runtime.getRuntime().addShutdownHook(new Thread(System.out::flush));
        }
    }

    static class Scanner {
        BufferedReader reader = new BufferedReader(new InputStreamReader(java.lang.System.in), 1 << 15);
        StringTokenizer tokenizer;

        String next() {
            try {
                while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                    tokenizer = new StringTokenizer(reader.readLine());
                }
            } catch (IOException ignored) {
            }
            return tokenizer.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        char[][] nextCharArray(int n, int m) {
            char[][] a = new char[n][m];
            for (int i = 0; i < n; i++) {
                a[i] = next().toCharArray();
            }
            return a;
        }

        int[] nextIntArray(int n, java.util.function.IntUnaryOperator op) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = op.applyAsInt(nextInt());
            }
            return a;
        }

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextLong();
            }
            return a;
        }

        long[] nextLongArray(int n, java.util.function.LongUnaryOperator op) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = op.applyAsLong(nextLong());
            }
            return a;
        }
    }
}