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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

string s;

int dfs(char c, unsigned int count_p, unsigned int count_g, int sum) {
  int judge, g;
  if (count_p + count_g > s.size())
    return sum;
  else {
    if (c == 'p') {
      if (s.at(count_p + count_g - 1) == 'p')
        judge = 0;
      else
        judge = 1;
    }
    else {
      if (s.at(count_p + count_g - 1) == 'p')
        judge = -1;
      else
        judge = 0;
    }
    g = dfs('g', count_p, count_g + 1, sum + judge);
    if (count_p < count_g)
      return max(dfs('p', count_p + 1, count_g, sum + judge), g);
    else
      return g;
  }
}

int main() {
  cin >> s;
  int count;
  count = dfs('g', 0, 1, 0);
  cout << count << endl;
}