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)

#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <random>

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;

typedef
tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> orderedSet;

#define pb push_back
#define F first
#define S second
#define all(a) (a).begin(), (a).end()

#define for1(i0, l0, r0) for (int i0 = l0; i0 < r0; ++i0)
#define for2(i0, l0, r0) for (int i0 = l0; i0 <= r0; ++i0)
#define forn(i0, n0) for (int i0 = 0; i0 < n0; ++i0)
#define forn1(i0, n0) for (int i0 = 1; i0 < n0; ++i0)
#define forr(i0, n0) for (int i0 = n0; i0 >= 0; --i0)
#define forr1(i0, r0, l0) for (int i0 = r0; i0 > l0; --i0)
#define forr2(i0, r0, l0) for (int i0 = r0; i0 >= l0; --i0)

#define Sort(a) sort(all(a))
#define Reverse(a) reverse(all(a))

typedef long double ld;
#define ui unsigned int
#define ull unsigned long long

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<string, string> pss;

typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<double> vd;
typedef vector<ld> vld;
typedef vector<bool> vb;
typedef vector<string> vs;

typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<pdd> vpdd;

typedef vector<vi> vvi;
typedef vector<vd> vvd;
typedef vector<vll> vvll;
typedef vector<vb> vvb;

#define vpss vector<pss>
#define vvs vector<vs>
#define vvpii vector<vpii>
#define vvpll vector<vpll>
#define vpt vector<pt>
#define vvvi vector<vvi>
#define vsi vector<set<int>>
#define ss second
#define ff first

#define printvi(arr) for (int x0 : arr) cout << x0 << ' '; cout << '\n';
#define printvll(arr) for (ll x0 : arr) cout << x0 << ' '; cout << '\n';
#define printpair(pair0) cout << pair0.F << ' ' << pair0.S << '\n';
#define printvp(arr) for (auto pair0 : arr) printpair(pair0);
#define initArray(arr, N0, X0) memset(arr, X0, N0 * sizeof(int))
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void init() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    //cout.precision(16);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
}

void solve() {
    string s;
    cin >> s;
    int n = s.length();
    int cnt = 0;
    for (char c : s){
        if (c == 'p') cnt++;
    }
    cout << n / 2 - cnt << '\n';
}

bool multitest = false;

int main() {
    init();
    int t = 1;
    if (multitest) cin >> t;
    forn(i, t) solve();
}