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>
using namespace std;
#define int long long
struct Fast {Fast(){std::cin.tie(0);ios::sync_with_stdio(false);}} fast;
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define F first
#define S second
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define RFOR(i, a, b) for (int i = 
#define REVERSE(x) reverse(ALL(x))
#define SORT(x) sort(ALL(x))
#define RSORT(x) sort(ALL(x), greater<>())
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))
#define SIZE(x) ll((x).size())
#define gcd __gcd
const int INF = 1001001001;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
template <typename T = int> inline T in(){
  T x; cin >> x; return x;}
template <typename T> inline void print(const T& x){
  cout << x << "\n";}
inline int lcm(int a, int b){
  return a / gcd(a, b) * b;}
bool is_prime(int n){
  if (n < 2) return false;
  FOR(i, 2, (int)(sqrt(n) + 1)){
    if (n % i == 0) return false;
  }
  return true;
}


signed main(){
  int N = in();
  string S = in<string>();
  
  int counter = 1;
  REP(i, N-1){
    if (S.at(i) != S.at(i+1)) ++counter;
  }
  print(counter);  
}