Problem:
Score : 300 points

Problem Statement

We have a string X, which has an even number of characters. Half the characters are S, and the other half are T.

Takahashi, who hates the string ST, will perform the following operation 10^{10000} times:

Among the occurrences of ST in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.

Find the eventual length of X.

Constraints

2 ≦ |X| ≦ 200,000

The length of X is even.

Half the characters in X are S, and the other half are T.

Partial Scores

In test cases worth 200 points, |X| ≦ 200.

Input

The input is given from Standard Input in the following format:

X

Output

Print the eventual length of X.

Sample Input 1

TSTTSS

Sample Output 1

4

In the 1-st operation, the 2-nd and 3-rd characters of TSTTSS are removed.
X becomes TTSS, and since it does not contain ST anymore, nothing is done in the remaining 10^{10000}-1 operations.
Thus, the answer is 4.

Sample Input 2

SSTTST

Sample Output 2

0

X will eventually become an empty string: SSTTST ⇒ STST ⇒ ST ⇒ ``.

Sample Input 3

TSSTTTSS

Sample Output 3

4

X will become: TSSTTTSS ⇒ TSTTSS ⇒ TTSS.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.LinkedList;

public class Main {

	static BufferedReader br;

	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));

		String s = readString();
		char[] ch = s.toCharArray();
		br.close();

		LinkedList<Character> ll = new LinkedList<Character>();
		int n = ch.length;
		ll.add(ch[0]);
		for (int i = 1; i < n; i++) {
			if (ch[i] == 'T') {
				if (ll.size() > 0) {
					char cp = ll.peekLast();
					if(cp=='S')ll.removeLast();
					else ll.add('T');
				

				}else ll.add('T');
			}else ll.add('S');
		}
		
		System.out.println(ll.size());

	}

	static public String readString() throws IOException {
		return br.readLine();

	}

	static public long readLong() throws IOException {
		return Long.parseLong(br.readLine());
	}

	static public int readInt() throws IOException {
		return Integer.parseInt(br.readLine());
	}

	static public int[] readIntArr() throws IOException {
		String[] str = br.readLine().split(" ");
		int arr[] = new int[str.length];
		for (int i = 0; i < arr.length; i++)
			arr[i] = Integer.parseInt(str[i]);
		return arr;
	}

	static public Integer[] readOIntArr() throws IOException {
		String[] str = br.readLine().split(" ");
		Integer arr[] = new Integer[str.length];
		for (int i = 0; i < arr.length; i++)
			arr[i] = new Integer(str[i]);
		return arr;
	}

	static public double[] readDoubleArr() throws IOException {
		String[] str = br.readLine().split(" ");
		double arr[] = new double[str.length];
		for (int i = 0; i < arr.length; i++)
			arr[i] = Double.parseDouble(str[i]);
		return arr;
	}

	static public Double[] readODoubleArr() throws IOException {
		String[] str = br.readLine().split(" ");
		Double arr[] = new Double[str.length];
		for (int i = 0; i < arr.length; i++)
			arr[i] = new Double(str[i]);
		return arr;
	}

	static public long[] readLongArr() throws IOException {
		String[] str = br.readLine().split(" ");
		long arr[] = new long[str.length];
		for (int i = 0; i < arr.length; i++)
			arr[i] = Long.parseLong(str[i]);
		return arr;
	}

	static public Long[] readOLongArr() throws IOException {
		String[] str = br.readLine().split(" ");
		Long arr[] = new Long[str.length];
		for (int i = 0; i < arr.length; i++)
			arr[i] = new Long(str[i]);
		return arr;
	}

	static public double readDouble() throws IOException {
		return Double.parseDouble(br.readLine());
	}
}