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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <string>
#include <cstring>

#define roop(i, n) for(int i=0;i<n;i++)

using namespace std;
using ll = long long;
using ull = unsigned long long;

constexpr ull mod = 1e9+7;
constexpr int N = 13;
ull dp[100010][N];

int myPower(int a, int n) { // aのn乗の計算
	int x = 1;
	while (n > 0) { // 全てのbitが捨てられるまで
		if ( n & 1) { // 一番右のbitが１のとき
			x *= a;
		}
		a *= a;
		n >>= 1; // bit全体を右に1つシフトして一番右を捨てる
	}
	return x;
}

int main() {
	int n, i;
	string s;
	cin >> n >> s;
    s.push_back('#');
	int ans = 0;
	for (i = 0; i + 1 < s.size(); i++) {
		if (s[i] != s[i + 1]) {
			ans++;
		}
	}
	cout << ans << endl;
	return 0;
}