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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.util.InputMismatchException;
import java.util.Stack;

public class Main {
	
	public static void main(String[] args) {
		Task task = new Task();
		task.solve();
	}

	public static class Task {
		static final long MOD = (long) 1e9 + 7;

		public void solve() {
			Scan sc = new Scan();
			Stack<Character> st = new Stack<Character>();
			char[] carr = sc.scanString().toCharArray();
			for(int i = 0, end = carr.length; i < end; ++i){
				char curr = carr[i];
				if(st.empty())
					st.add(curr);
				else {
					if (st.peek() == 'S' && curr == 'T')
						st.pop();
					else
						st.add(curr);
				}
			}
			System.out.println(st.size());
		}
	}

	public static class Scan {
		private byte[] buf = new byte[1024];
		private int index;
		private InputStream in;
		private int total;

		public Scan() {
			in = System.in;
		}

		public int scan() {
			if (total < 0)
				throw new InputMismatchException();
			if (index >= total) {
				index = 0;
				try {
					total = in.read(buf);
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
				if (total <= 0)
					return -1;
			}
			return buf[index++];
		}

		public long scanLong() {
			long integer = 0;
			int n = scan();
			while (isWhiteSpace(n))
				n = scan();
			int neg = 1;
			if (n == '-') {
				neg = -1;
				n = scan();
			}
			while (!isWhiteSpace(n)) {
				if (n >= '0' && n <= '9') {
					integer *= 10;
					integer += n - '0';
					n = scan();
				} else
					throw new InputMismatchException();
			}
			return neg * integer;
		}

		public double scanDouble() throws IOException {
			double doub = 0;
			int n = scan();
			while (isWhiteSpace(n))
				n = scan();
			int neg = 1;
			if (n == '-') {
				neg = -1;
				n = scan();
			}
			while (!isWhiteSpace(n) && n != '.') {
				if (n >= '0' && n <= '9') {
					doub *= 10;
					doub += n - '0';
					n = scan();
				} else
					throw new InputMismatchException();
			}
			if (n == '.') {
				n = scan();
				double temp = 1;
				while (!isWhiteSpace(n)) {
					if (n >= '0' && n <= '9') {
						temp /= 10;
						doub += (n - '0') * temp;
						n = scan();
					} else
						throw new InputMismatchException();
				}
			}
			return doub * neg;
		}

		public String scanString() {
			StringBuilder sb = new StringBuilder();
			int n = scan();
			while (isWhiteSpace(n))
				n = scan();
			while (!isWhiteSpace(n)) {
				sb.append((char) n);
				n = scan();
			}
			return sb.toString();
		}

		public char scanChar() {
			int n = scan();
			while (isWhiteSpace(n))
				n = scan();
			return (char) n;
		}

		private boolean isWhiteSpace(int n) {
			if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1)
				return true;
			return false;
		}
	}
}