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 <sstream>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <deque>
#include <stack>
#include <iomanip>
#include <cmath>


using namespace std;


typedef long long ll;
typedef long double ld;


const ll MOD = 1000000007ll;


ll modSum(ll a, ll b) {
  return ((a % MOD) + (b % MOD)) % MOD;
}


ll modSubtract(ll a, ll b) {
  return (((a % MOD) - (b % MOD)) + MOD + MOD) % MOD;
}


ll modProd(ll a, ll b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}


ll getPowMod(ll x, ll e) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    ll tmp = getPowMod(x, e/2);
    return modProd(tmp, tmp);
  } else {
    ll tmp = getPowMod(x, e-1);
    return modProd(tmp, x);
  }
}


ll getPow(ll x, ll e) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    ll tmp = getPow(x, e/2);
    return tmp * tmp;
  } else {
    ll tmp = getPow(x, e-1);
    return tmp * x;
  }
}


ll getInverse(ll x) {
  return getPowMod(x, MOD-2);
}


bool isEven(ll x) {
  ll tmp = ((x % 2) + 2) % 2;
  return tmp == 0;
}


ll getSumOfDigitsInBase(ll n, ll b) {
  ll ret = 0;
  while (n > 0) {
    ret += n % b;
    n /= b;
  }
  return ret;
}


vector<int> getKMP(string &s) {
  int len = (int)s.size();
  vector<int> ret (len, 0);
  for (int i = 1 ; i < len ; i++) {
    int at = ret[i-1];
    while (at > 0 && s[i] != s[at]) {
      at = ret[at-1];
    }
    if (s[i] == s[at]) {
      at++;
    }
    ret[i] = at;
  }
  return ret;
}


string getSubstring(string &s, int from, int to) {
  int l = to-from+1;
  if (l <= 0) {
    return "";
  }
  return s.substr(from, l);
}

ll gcd(ll a, ll b, ll & x, ll & y) {
    if (a == 0) {
        x = 0;
        y = 1;
        return b;
    }
    ll x1, y1;
    ll d = gcd(b % a, a, x1, y1);
    x = y1 - (b / a) * x1;
    y = x1;
    return d;
}














string s;


void input() {
  cin >> s;
}


void preprocess() {
}


void solve() {
  int ans = 0;
  int cnt = 0;
  for (char ch : s) {
    if (ch == 'g') {
      if (cnt > 0) {
        ans++;
        cnt--;
      } else {
        cnt++;
      }
    } else {
      if (cnt > 0) {
        cnt--;
      } else {
        ans--;
        cnt++;
      }
    }
  }
  cout << ans;
}


int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);

  int T;
//  cin >> T;
  T = 1;

  for (int i = 0 ; i < T ; i++) {
    input();
    preprocess();
    solve();
  }

  return 0;
}