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>
#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<algorithm>
#include<map>
using namespace std;
#define rep(i,x) for(int i = 0; i < x; i++)
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
typedef pair<ll,ll> Pll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
const ll MOD = 1e9+7;
const int n_max = 1e5+10;

int main(){
    ll n; cin >> n;
    string s;cin >> s;
    ll res = 1;
    rep(i,n){
        if(!i)continue;
        if(s[i] != s[i-1])res++;
    }
    cout << res << endl;
    return 0;
}