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>
using namespace std;
// make pair, tuple
// __gcd(int, int)
// bool isPrime(int n){
//       if(n <= 1)return false;
//       if(n <= 3)return true;
//       if(n % 2 == 0 || n % 3 == 0)return false;
//       for(int i = 5;i * i <= n;i += 6){
//             if(n % i == 0 || n % (i + 2) == 0)return false;
//       }
//       return true;
// }
// int countSubString(string s, string temp){
//       int ans = 0;
//       size_t f = s.find(temp);
//       while(f != string::npos){
//             ans++;
//             f = s.find(temp, f + 1);
//       }
//       return ans;
// }

int main(){
    int n;cin >> n;
    string s;cin >> s;
    int ans = 1;
    for(int i = 1;i < n;i++){
        if(s[i] != s[i - 1])ans++;
    }
    cout << ans;
}