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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using ll = long long;
using namespace std;
#define rep(i,n) for(int i=0, i##_len=(int)(n); i<i##_len; i++)
#define reps(i,n) for(int i=1 , i##_len=(int)(n);i<=i##_len;i++)
#define rrep(i,n) for(int i=((int)(n)-1);i>=0;i--)
#define rreps(i,n) for(int i=((int)(n));i>0;i--)
#define repi(i,x) for(auto i=(x).begin(),i##_fin=(x).end();i!=i##_fin;i++)
#define all(x) (x).begin(), (x).end()
#define F first
#define S second
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back
string solve(bool a) { return ((a) ? "Yes" : "No"); }
typedef vector<int> Vi;
typedef vector<Vi> VVi;
typedef pair<int , int> Pi;
typedef vector<Pi> VPi;
typedef vector<long long> V;
typedef vector<V> VV;
typedef pair<long long , long long> P;
typedef vector<P> VP;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1;} return 0;}
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1;} return 0;}
template <class T, class U>ostream& operator<<(ostream& os, const pair<T, U>& p) {
	os << "(" << p.first << "," << p.second << ")";
	return os;
}
template <class T>ostream& operator<<(ostream& os, const vector<T>& v) {
	os << "{";
	rep(i, v.size()) {
		if (i) os << ",";
		os << v[i];
	}
	os << "}";
	return os;
}
template <class T, class U>istream& operator>>(istream& is, pair<T, U>& p) {
	is >> p.first >>p.second ;
	return is;
}
template <class T>istream& operator>>(istream& is,  vector<T>& v) {
	rep(i, v.size()) {
		is >> v[i];
	}
	return is;
}
const long long INFLL = 1LL<<60;
const int INF = 1<<30;
const double PI=acos(-1);
int main(){
	string s;
	int ans=0;
	cin >>s;
	int p = 0;
	Pi pg = mp(0,0);
	for (char i:s){
		p += (i == 'p');
	} 
	for(char i:s){
		if(i=='p'){
			pg.F++;
			clog << 'p';
		}else{
			if(pg.S <p or pg.F == pg.S){
				pg.S++;
				clog << 'g';
			}else{
				ans++;
				pg.F++;
				p++;
				clog << 'p';
			}
		}
	}

	cout << ans << endl;
}