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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h> 
 
using namespace std;
 
 template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//typedef
//------------------------------------------
typedef pair<int, int> PII;
typedef pair<long, long> PLL;
typedef long long LL;
//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define PB emplace_back
#define MP make_pair 
#define SZ(a) int((a).size())
//repetition
//------------------------------------------
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define SORT(c) sort((c).begin(),(c).end())
//constant
//--------------------------------------------
//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))
 

int main(){
    string S;
    cin>>S;
    int g=0,p=0;
    int win=0,lose=0;
    REP(i,SZ(S)){
        if(S[i]=='g'){
            if(p<=g){
                win++;p++;
            }else{
                g++;
            }
        }else{
            if(p<=g){
                p++;
            }else{
                lose++;g++;
            }
        }
    }
    cout<<win-lose<<endl;
    return 0;
}