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 <algorithm> // std::sort(begin,end), std::min, std::max
#include <iostream>
#include <queue>
#include <string>
#include <unordered_map>
#include <utility> // std::swap
#include <vector>

using namespace std;
using ll = long long;
using pii = pair<int, int>;

template <typename T> // 0<1<2<3<4
using pq_asc = priority_queue<T, vector<T>, greater<T>>;
template <typename T> // 0>1>2>3>4
using pq_dsc = priority_queue<T>;

int main(int argc, char const *argv[]) {
  ll n;
  string s;
  cin >> n >> s;
  int ret = 0;
  char prev_c = ' ';
  for (char c : s) {
    if (prev_c != c) {
      ret++;
    }
    prev_c = c;
  }

  cout << ret << endl;

  return 0;
}