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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;

public class Main {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		char[] s = sc.next().toCharArray();

		int pt = 0;
		for (int i = 0; i < s.length; i++) {
			if (i % 2 == 1 && s[i] == 'g')
				pt++;
			else if (i % 2 == 0 && s[i] == 'p')
				pt--;
		}
		System.out.println(pt);
	}

}

class FastScanner {
	public static String debug = null;

	private final InputStream in = System.in;
	private int ptr = 0;
	private int buflen = 0;
	private byte[] buffer = new byte[1024];
	private boolean eos = false;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				if (debug != null) {
					buflen = debug.length();
					buffer = debug.getBytes();
					debug = "";
					eos = true;
				} else {
					buflen = in.read(buffer);
				}
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen < 0) {
				eos = true;
				return false;
			} else if (buflen == 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	private void skipUnprintable() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
	}

	public boolean isEOS() {
		return this.eos;
	}

	public boolean hasNext() {
		skipUnprintable();
		return hasNextByte();
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	public int nextInt() {
		return (int) nextLong();
	}

	public long[] nextLongList(int n) {
		return nextLongTable(1, n)[0];
	}

	public int[] nextIntList(int n) {
		return nextIntTable(1, n)[0];
	}

	public long[][] nextLongTable(int n, int m) {
		long[][] ret = new long[n][m];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				ret[i][j] = nextLong();
			}
		}
		return ret;
	}

	public int[][] nextIntTable(int n, int m) {
		int[][] ret = new int[n][m];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				ret[i][j] = nextInt();
			}
		}
		return ret;
	}
}