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: Time Limit Exceeded)

#include<bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < (int)n; i++)
#define MOD 1000000009
using namespace std;
typedef long long ll;
int main()
{
  int n;
  string s;
  cin >> n >> s;
  int counter = 0;
  char tmp;
  REP(i,n){
    tmp = s[i];
//	cout << i << "  " << tmp << endl;
    for(int j = i + 1; j < n; j ++){
      if(s[j] == tmp){
	  
      }else{
		counter ++;
		i = j - 1;
		break;
      }
    }
  }
  cout << counter + 1 << endl;
}