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)

#include<iostream>
#include<map>
#include<set>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef int itn;

const int inf = 99999999;
const ll linf = 999999999999999999;
const double pi = acos(-1);
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};

#define p_queue(i) priority_queue<i> 
#define rp_queue(i) priority_queue<i, vector<i>, greater<i>> 
#define P(p) cout<<(p)<<endl
#define rep(i,m,n) for(int i = (m); i < (int)(n); i++)
#define mod(i) ((i)%(ll)(1e9 + 7))
#define divm(a,b) (mod(a * modpow((ll)b,(ll)(1e9+5))))
#define rsort(a,b,c) sort(a, b, greater<c>())
#define vsort(v) sort(v.begin(), v.end());
#define rvsort(v) sort(v.end(), v.begin());
#define ft first
#define sd second
#define pb push_back
#define it insert
#define sz(x) ((int)(x).size())
#define lb(a,n,k) (lower_bound(a,a+n,k) - a) 
#define vlb(a,k)  (lower_bound(a.begin(),a.end(),k) - a.begin())
#define YES cout<<"YES"<< endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl  
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl
#define ret return

ll modpow(ll i,ll j){ ll tmp=1; while(j){ if(j%2)tmp=mod(tmp*i);i=mod(i*i);j/=2;}return tmp;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

//ll ncr[100][100];
//ll nCr(int n, int r){if(n==r) ret ncr[n][r] = 1; if(r==0) ret ncr[n][r] = 1; if(r==1) ret ncr[n][r] = n;if(ncr[n][r]) ret ncr[n][r]; ret ncr[n][r] = nCr(n-1,r) + nCr(n-1,r-1);}

//ll npr[100][100]={};
//ll nPr(int n,int r){if(npr[n][r])ret npr[n][r];if(r==0)ret npr[n][r] = 1;if(r==1)ret npr[n][r] = n;ret npr[n][r] = n * nPr(n-1,r-1);}

//ll nHr(int n,int r){ret nCr(n+r-1,r);}

///////////////////////////////////////////////////////////////////////////



int main(){
	string s;
	cin >> s;

    ll ans = 0,st = 0;
	rep(i,0,sz(s)){
	    if(st > 0){
            if(s[i] == 'p'){
                st--;
            }else{
                st--;
                ans++;
            }
	    }else{
            if(s[i] == 'p'){
                ans --;
                st++;
              //  cout << 'g';
            }else{
                st++;
            }
	    }
	}
    P(ans);
	return  0;
}