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: Accepted)

/*input
ggppgggpgggpg

*/
#include <bits/stdc++.h>
#include <stdlib.h>
#include <time.h> 

using namespace std;
long long mod=1e9+7;
typedef long long int lld;
#define rep(i,a,n) for(long long int i = (a); i <= (n); ++i)
#define repI(i,a,n) for(int i = (a); i <= (n); ++i)
#define repD(i,a,n) for(long long int i = (a); i >= (n); --i)
#define repDI(i,a,n) for(int i = (a); i >= (n); --i)
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define sc(a) scanf("%lld",&a)
#define sc2(a,b) scanf("%lld%lld",&a,&b)
#define sc3(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)
#define scd(a) scanf("%d",&a)
#define scd2(a,b) scanf("%d%d",&a,&b)
#define scd3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scf(a) scanf("%lf",&a)
#define scf2(a,b) scanf("%lf%lf",&a,&b)
#define scf3(a,b,c) scanf("%lf%lf%lf",&a,&b,&c)
#define prL(a) printf("%lld\n",a)
#define prS(a) printf("%lld ",a)
#define prdL(a) printf("%d\n",a)
#define prdS(a) printf("%d ",a)
#define Error(x) cerr << #x << " = " << (x) <<endl;
#define Error2(a,b) cerr<<"( "<<#a<<" , "<<#b<<" ) = ( "<<(a)<<" , "<<(b)<<" )\n";
#define Error3(a,b,c) cerr<<"("<<#a<<" , "<<#b<<" , "<<#c<<" ) = ( "<<(a)<<" , "<<(b)<<" , "<<(c)<<")\n";
#define Error4(a,b,c,d) cerr<<"("<<#a<<" , "<<#b<<" , "<<#c<<" , "<<#d<<" ) = ( "<<(a)<<" , "<<(b)<<" , "<<(c)<<" , "<<(d)<<")\n";
#define errop(a) cerr<<#a<<" = ( "<<((a).x)<<" , "<<((a).y)<<" )\n";
#define popcount __builtin_popcountll
typedef pair<lld,lld> PA;
#define lim 1000010
#define lim2 3003
// inline lld sqr(lld x) { return x * x; }
// map<lld,lld>::iterator it;
// std::ios::sync_with_stdio(false);
// priority_queue<PA> Q;
// lld dp[1<<18];

lld B[lim],A[lim];
lld n,m,p,q;
string S[lim];

lld powP(lld a,lld b){
    if(b==0) return 1%mod;
    lld k;
    k=powP(a,b/2);
    k=k*k%mod;
    if(b%2==0) return k;
    else return a*k%mod;
}
bool bitSet(lld n,lld i){
    if((n&(1<<i))!=0) return true;
    else return false;
}

void mmod(long long int& cur) {
    while (cur >= mod) {
        cur -= mod;
    }
}


int main(){
    lld T,i,j,h,l,r,k,s,b,c,a,d,w,x,y,v,z,t,curr,prev,sum,ans,pos,val,countA,secondMin,indicator;
    string S;
    cin>>S;
    n=S.size();
    lld score=0;
    a=b=0;
    rep(i,0,n-1){
        if(a<b){
            a++;
            t=1;
        }
        else{
            b++;
            t=0;
        }
        if(S[i]=='p'){
            if(t==0) score--;
        }
        else{
            if(t==1) score++;
        }
    }
    prL(score);
    return 0;
}