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>
// Begin Header {{{
#define all(x) (x).begin(), (x).end()
#define lli long long int
#define rep(i,n) for(lli i=0; i<n; i++)
#define loop(i, j, n) for(lli i=j; i<n; i++)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
constexpr int INF  = 0x3f3f3f3f;
const long long mod=1e9+7;
//Function
lli combi(lli n, lli r)
{
	if(r==0 || r==n) return 1;
	else return combi(n-1,r) + combi(n-1,r-1);
}
lli sigma1(lli n){return n*(n+1)/2;}
lli sigma2(lli s, lli n){return sigma1(n) - sigma1(s -1);}
using namespace std;
// }}} End Header

int main()
{
	ios::sync_with_stdio(false); cin.tie(nullptr);
	lli n, cnt=1;
	string s;
	cin >> n >> s;
	rep(i,n-1){
		if(s[i] != s[i+1]) cnt++;
	}
	cout << cnt << endl;
	return 0;
}