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>
#define sz(arr) (int)(arr).size()
#define rng(arr) arr.begin(), arr.end()
#define show(x) cout << #x << " = " << x << endl;

using namespace std;
typedef long long int ll;

void _cin() {}
template <class Head, class... Tail>
void _cin(Head&& head, Tail&&... tail) {
  cin >> head;
  _cin(forward<Tail>(tail)...);
}

#define Cin(T, ...) \
  T __VA_ARGS__;    \
  _cin(__VA_ARGS__)

#define Cins(T, n, xs) \
  vector<T> xs(n);     \
  for (int i = 0; i < n; ++i) cin >> xs[i]

void Main();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(15);
  Main();
  return 0;
}

void Main() {
  Cin(int, n);
  Cin(string, s);

  int ans = 1;
  for (int i = 1; i < n; ++i) {
    if (s[i] != s[i - 1]) ans++;
  }

  cout << ans << endl;
}