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: Runtime Error)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<tuple>
#include<map>
#include<queue>
#define rep(i, n) rep2(i,0,n)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
typedef long long ll;
using namespace std;

int main() {
	int n;
	cin >> n;
	int sum = 1;
	vector<char> s(n);
	rep(i, n) cin >> s.at(i);
	int i = 0;
	while (i<n) {
		while (s.at(i) == s.at(i + 1)&&i<n-1) {
			++i;
		}
		i++;
		if (i == n-1) {
			sum++;
			break;
		}
		
		sum++;
		
	}
	cout << sum << endl;
}