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 debug(x) cerr << #x << ": " << (x) << endl
using namespace std;

using ull = unsigned long long;
using ll = long long;

using Int = int;
using Double = double;

using P = pair<Int, Int>;
using V = vector<Int>;
using M = vector<V>;

const Double PI  = 3.141592653589793238463;
const Int MOD = 1000000007;
const Int INF = numeric_limits<Int>::max() / 3;

template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
    for(T& x: vec){
        is >> x;
    }
    return is;
}

Int solve(Int n, string& s){
    Int m = 0;
    char x = 'X';
    for(char c: s){
        if(c != x){
            x = c;
            m++;
        }
    }
    return m;
}

int main(int argc, char* argv[]){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);

    Int n;
    string s;
    cin >> n >> s;
    cout << solve(n, s) << "\n";

    return 0;
}