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 <bits/stdc++.h>
#define ll long long
using namespace std;

string s;

int n;

int pr[111111], mn[111111];

int main() {
#ifdef LOCAL
    freopen("xxx.in", "r", stdin);
    freopen("xxx.out", "w", stdout);
#endif
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> s;
    n = s.size();
    for (int i = 1; i <= n; ++i) {
        pr[i] = pr[i - 1] + (s[i - 1] == 'p');
    }
    mn[n - 1] = (n - 2 * pr[n]) / 2;
    for (int i = n - 2; i >= 0; --i) {
        mn[i] = min(mn[i + 1], (i + 1 - 2 * pr[i + 1]) / 2);
    }
    int cnt = 0;
    for (int i = 0; i < n; ++i) {
        if (s[i] == 'g') {
            if (mn[i] - cnt > 0)
                cnt++;
        }
    }
    cout << cnt;
    return 0;
}