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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rgi register int
#define ll long long
#define il inline

using namespace std;

int n, len, ans;
string str;

il int read() {
	rgi x = 0, f = 0, ch;
	while(!isdigit(ch = getchar()))	f |= ch == '-';
	while(isdigit(ch))	x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
	return f ? -x : x;
}

int main() {
	n = read();
	cin >> str; len = str.length(); ans = len;
	for(rgi i = 0; i < len - 1; ++i) {
		if(str[i + 1] == str[i])
			ans--;
	}
	printf("%d", ans);
	return 0;
}