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;
#define int long long
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
using P=pair<long long,long long>;
#define rep(i,n) for(long long i=0; i<(long long)n; i++)
#define FOR(i,a,b) for(long long i=a; i<b; i++)
#define ll long long
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((long long)(x).size())
#define COUT(x) cout << x << endl
#define PB(x) push_back(x)
#define MP make_pair
#define vint vector<int>
#define vvint vector<vector<int>>
#define vstr vector<string>
#define vp vector<pair<int,int>>
#define SUM(x) accumulate(x.begin(),x.end(),0)
#define MAX(x) *max_element(x.begin(),x.end())
#define MIN(x) *min_element(x.begin(),x.end())
#define couty cout<<'Y'<<'e'<<'s'<<endl
#define coutn cout<<'N'<<'o'<<endl
#define coutY cout<<'Y'<<'E'<<'S'<<endl
#define coutN cout<<'N'<<'O'<<endl
#define yn(x) cout<<(x?"Yes":"No")<<endl
#define YN(x) cout<<(x?"YES":"NO")<<endl
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}
long long lcm(long long a, long long b){return a*b/gcd(a,b);}
const long long dx[4]={1,0,-1,0};
const long long dy[4]={0,1,0,-1};
const long long INF = 1e12;
const long long MOD = 1e9+7;
//cout<<fixed<<setprecision(15);

signed main(){
    int n; cin >> n;
    string s; cin >> s;
    int res=1;
    rep(i,n-1){
        if(s[i]!=s[i+1]) res++;
    }
    cout<<res<<endl;
}