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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <climits>
#include <cmath>
#include <unordered_set>
//#define local
#ifdef local
#include "dbg-macro/dbg.h"
#endif
#define ll long long
#define ull unsigned long long
#define pi (acos(-1))
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for (unsigned long long i = 0; i < (unsigned long long)(n); ++i)
#define vec std::vector
using namespace std;
std::vector<unsigned ll> genprimevec(const unsigned ll N);
int main()
{
    int n;
    vec<char> s;
    cin >> n ;
    char t;
    cin>>t;
    s.push_back(t);
    rep(i, n - 1)
    {
        cin >> t;
        if (s[s.size()-1] != t) { s.push_back(t); }
    }
    cout<<s.size()<<endl;
    return 0;
}

ll extgcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1, y = 0;
        return a;
    }
    ll d = extgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}


std::vector<unsigned ll> genprimevec(const unsigned ll N)
{
    std::vector<bool> is_prime(N + 1);
    for (unsigned ll i = 0; i <= N; i++) { is_prime[i] = true; }
    std::vector<unsigned ll> P;
    for (unsigned ll i = 2; i <= N; i++)
    {
        if (is_prime[i])
        {
            for (unsigned ll j = 2 * i; j <= N; j += i) { is_prime[j] = false; }
            P.emplace_back(i);
        }
    }
    return P;
}