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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <set>
#include <map>
#include <algorithm>

#define int long long
#define MOD7 1000000007
#define MOD9 1000000009

#define rep(i, n) for (int i = 0; i < (n); i++)
#define all(a) (a).begin(), (a).end()

using namespace std;

int nextInt() {int a; cin >> a; return a;}
char nextChar() {char a; cin >> a; return a;}
double nextDouble() {double a; cin >> a; return a;}
string nextString() {string a; cin >> a; return a;}

void inputVector(vector<int> &v, int &n) {rep(i,n){v.push_back(nextInt());}}
void inputVector(vector<double> &v, int &n) {rep(i,n){v.push_back(nextDouble());}}
void inputVector(vector<string> &v, int &n) {rep(i,n){v.push_back(nextString());}}

int cnt[2];

signed main() {
    string S;
    cin >> S;

    rep(i, S.size()) {
	if (S[i] == 'p') {
	    cnt[0]++;
	} else {
	    cnt[1]++;
	}
    }

    cout << (cnt[1] - cnt[0]) / 2 << endl;

    return 0;
}