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 Ruby (2.3.3) (judge verdict: Accepted)

def get_i() #空白区切の入力を数値(整数)の配列で返す
  return gets.chomp.split(" ").map(&:to_i)
end
def get_f() #空白区切の入力を数値(実数)の配列で返す
  return gets.chomp.split(" ").map(&:to_f)
end
def get() #空白区切の入力を文字列の配列で返す
  return gets.chomp.split(" ")
end
def get_nsp() #入力されたものを一文字ずつに区切った文字列の配列で返す
  return gets.chomp.split("")
end
def yn_judge(bool,y="Yes",n="No") #boolに真偽を投げることで、trueならy、falseならnの値を出力する
  return bool ? y : n 
end
def array(size1,init=nil,size2=1) #size2に二次元配列時の最初の要素数、size1に次の配列の大きさ、initに初期値を投げることでその配列を返す
  if size2==1
    return Array.new(size1){init}
  else
    return Array.new(size2){Array.new(size1){init}}
  end
end

N=gets.to_i
s=get_nsp
ans=1
(N-1).times do|i|
    if s[i]!=s[i+1]
      ans+=1 
    end
end
puts ans