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>
using namespace std;

// Define
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <class T> using pvector = vector<pair<T, T>>;
template <class T>
using rpriority_queue = priority_queue<T, vector<T>, greater<T>>;
constexpr const ll dx[4] = {1, 0, -1, 0};
constexpr const ll dy[4] = {0, 1, 0, -1};
constexpr const ll MOD = 1e9 + 7;
constexpr const ll mod = 998244353;
constexpr const ll INF = 1LL << 60;
constexpr const ll inf = 1 << 30;
constexpr const char rt = '\n';
constexpr const char sp = ' ';
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back
#define elif else if
#define all(a, v, ...)                                                         \
    ([&](decltype((v)) w) { return (a)(begin(w), end(w), ##__VA_ARGS__); })(v)
#define rall(a, v, ...)                                                        \
    ([&](decltype((v)) w) { return (a)(rbegin(w), rend(w), ##__VA_ARGS__); })(v)
#define fi first
#define se second

template <class T> bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> bool chmin(T &a, const T &b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}

// Debug
#define debug(...)                                                             \
    {                                                                          \
        cerr << __LINE__ << ": " << #__VA_ARGS__ << " = ";                     \
        for (auto &&X : {__VA_ARGS__}) cerr << "[" << X << "] ";               \
        cerr << rt;                                                            \
    }

#define dump(a, h, w)                                                          \
    {                                                                          \
        cerr << __LINE__ << ": " << #a << " = [" << rt;                        \
        rep(_i, h) {                                                           \
            rep(_j, w) cerr << a[_i][_j] << sp;                                \
            cerr << rt;                                                        \
        }                                                                      \
        cerr << "]" << rt;                                                     \
    }

#define vdump(a, n)                                                            \
    {                                                                          \
        cerr << __LINE__ << ": " << #a << " = [";                              \
        rep(_i, n) if (_i) cerr << sp << a[_i];                                \
        else cerr << a[_i];                                                    \
        cerr << "]" << rt;                                                     \
    }

// Loop
#define inc(i, a, n) for (ll i = (a), _##i = (n); i <= _##i; ++i)
#define dec(i, a, n) for (ll i = (a), _##i = (n); i >= _##i; --i)
#define rep(i, n) for (ll i = 0, _##i = (n); i < _##i; ++i)
#define each(i, a) for (auto &&i : a)

// Stream
#define fout(n) cout << fixed << setprecision(n)
struct io {
    io() { cin.tie(nullptr), ios::sync_with_stdio(false); }
} io;

// Speed
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

// Math
inline constexpr ll gcd(const ll a, const ll b) {
    return b ? gcd(b, a % b) : a;
}
inline constexpr ll lcm(const ll a, const ll b) { return a / gcd(a, b) * b; }

inline constexpr ll modulo(const ll n, const ll m = MOD) {
    ll k = n % m;
    return k + m * (k < 0);
}
inline constexpr ll chmod(ll &n, const ll m = MOD) {
    n %= m;
    return n += m * (n < 0);
}
inline constexpr ll mpow(ll a, ll n, const ll m = MOD) {
    ll r = 1;
    rep(i, 64) {
        if (n & (1LL << i)) r *= a;
        chmod(r, m);
        a *= a;
        chmod(a, m);
    }
    return r;
}
inline ll inv(const ll n, const ll m = MOD) {
    ll a = n, b = m, x = 1, y = 0;
    while (b) {
        ll t = a / b;
        a -= t * b;
        swap(a, b);
        x -= t * y;
        swap(x, y);
    }
    return modulo(x, m);
}

signed main() {
    string S;
    cin >> S;
    cout << (all(count, S, 'g') - all(count, S, 'p')) / 2 << rt;
}

// -g -D_GLIBCXX_DEBUG -fsanitize=undefined