Problem:
Score : 300 points

Problem Statement

AtCoDeer the deer and his friend TopCoDeer is playing a game.
The game consists of N turns.
In each turn, each player plays one of the two gestures, Rock and Paper, as in Rock-paper-scissors, under the following condition:

(※) After each turn, (the number of times the player has played Paper)≦(the number of times the player has played Rock).

Each player's score is calculated by (the number of turns where the player wins) - (the number of turns where the player loses), where the outcome of each turn is determined by the rules of Rock-paper-scissors.

(For those who are not familiar with Rock-paper-scissors: If one player plays Rock and the other plays Paper, the latter player will win and the former player will lose. If both players play the same gesture, the round is a tie and neither player will win nor lose.)

With his supernatural power, AtCoDeer was able to foresee the gesture that TopCoDeer will play in each of the N turns, before the game starts.
Plan AtCoDeer's gesture in each turn to maximize AtCoDeer's score.

The gesture that TopCoDeer will play in each turn is given by a string s. If the i-th (1≦i≦N) character in s is g, TopCoDeer will play Rock in the i-th turn. Similarly, if the i-th (1≦i≦N) character of s in p, TopCoDeer will play Paper in the i-th turn.

Constraints

1≦N≦10^5

N=|s|

Each character in s is g or p.

The gestures represented by s satisfy the condition (※).

Input

The input is given from Standard Input in the following format:

s

Output

Print the AtCoDeer's maximum possible score.

Sample Input 1

gpg

Sample Output 1

0

Playing the same gesture as the opponent in each turn results in the score of 0, which is the maximum possible score.

Sample Input 2

ggppgggpgg

Sample Output 2

2

For example, consider playing gestures in the following order: Rock, Paper, Rock, Paper, Rock, Rock, Paper, Paper, Rock, Paper. This strategy earns three victories and suffers one defeat, resulting in the score of 2, which is the maximum possible score.

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

--相手と同じ手を出せば必ず引き分けにはなる。
--p,gを出すトータルが同じとき、出すタイミングによって点数が変わるかどうかについては、
--これはかわらない。こちらの出す手の任意の二箇所を入れ替えても、点数の変動は相殺してゼロ。
--従って、p,gを出す回数を固定したら、それは履歴に関する出し手制約を満たしていればいつ出しても点数はかわらない。
--つまり勝てるときに勝っておいて、そのせいで後で負けても、トータルに支障はない。
--ということで貪欲法で行く。相手の出方を見ながら、こちらがより高い点を獲得できる手が出せるならそれを出す。


main::IO()
main = do
  enemy<-getLine
  print (solver enemy)

solver ::String -> Int
solver cs = let (_,_,score) = foldl decide_result (0,0,0) cs in score

decide_result:: (Int,Int,Int) -> Char -> (Int,Int,Int)
decide_result (p,g,score) hand
          |(hand=='g')&&(p <g) = (p+1,g,score+1)
          |(hand=='g')&&(p>=g) = (p,g+1,score)
          |(hand=='p')&&(p <g) = (p+1,g,score)
          |(hand=='p')&&(p>=g) = (p,g+1,score-1)