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.util.Arrays;
import java.util.Scanner;

public class Main {
    static final Scanner sc  = new Scanner(System.in);
    static final int     MOD = (int) 1E9 + 7;
    
    public static void main(String[] args) {
        int N = nint();
        char[] S = nchars();
        char cur = S[0];
        int count = 1;
        for (int i = 1; i < N; i++) {
            if (S[i] != cur) {
                count++;
                cur = S[i];
            }
        }
        System.out.println(count);
    }
    
    private static int nint() {
        return sc.nextInt();
    }

    private static int[] nints(int n) {
        return nints(n, 0, 0);
    }

    private static int[] nints(int n, int padL, int padR) {
        int[] a = new int[padL + n + padR];
        for (int i = 0; i < n; i++)
            a[padL + i] = nint();
        return a;
    }

    private static long nlong() {
        return sc.nextLong();
    }

    private static long[] nlongs(int n) {
        return nlongs(n, 0, 0);
    }

    private static long[] nlongs(int n, int padL, int padR) {
        long[] a = new long[padL + n + padR];
        for (int i = 0; i < n; i++)
            a[padL + i] = nlong();
        return a;
    }

    private static double ndouble() {
        return sc.nextDouble();
    }

    private static double[] ndoubles(int n) {
        return ndoubles(n, 0, 0);
    }

    private static double[] ndoubles(int n, int padL, int padR) {
        double[] d = new double[n + padL + padR];
        for (int i = 0; i < n; i++) {
            d[padL + i] = ndouble();
        }
        return d;
    }

    private static String nstr() {
        return sc.next();
    }

    private static char[] nchars() {
        return sc.next().toCharArray();
    }

    private static char[] nchars(int padL, int padR) {
        char[] temp = sc.next().toCharArray();
        char[] ret = new char[temp.length + padL + padR];
        System.arraycopy(temp, 0, ret, padL, temp.length);
        return ret;
    }

    private static char[][] nchars2(int h, int w) {
        return nchars2(h, w, 0, 0);
    }

    private static char[][] nchars2(int h, int w, int padLU, int padRD) {
        char[][] a2 = new char[h + padLU + padRD][w + padLU + padRD];
        for (int i = 0; i < h; i++)
            System.arraycopy(nchars(), 0, a2[padLU + i], padLU, w);
        return a2;
    }

    private static long min(long... ls) {
        return Arrays.stream(ls).min().getAsLong();
    }

    private static int min(int... is) {
        return Arrays.stream(is).min().getAsInt();
    }

    private static long max(long... ls) {
        return Arrays.stream(ls).max().getAsLong();
    }

    private static int max(int... is) {
        return Arrays.stream(is).max().getAsInt();
    }

    private static long abs(long a, long b) {
        return Math.abs(a - b);
    }
}