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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.NoSuchElementException;
import java.util.Set;

public class Main implements Runnable {
    public static void main(String[] args) {
        new Thread(null,new Main(), "" ,Runtime.getRuntime().maxMemory()).start();
    }

    // メイン処理
    public void run() {
        FastScanner sc = new FastScanner();
        char[] x = sc.next().toCharArray();
        int[] countS = new int[x.length];
        int[] countT = new int[x.length];

        if(x[0] == 'S') countS[0] = 1;
        else countT[0] = 1;
        for(int i = 1;i < x.length;i++) {
            if(x[i] == 'S') {
                countS[i] = countS[i - 1] + 1;
                countT[i] = countT[i - 1];
            }
            else {
                if(countS[i - 1] > 0) {
                    countS[i] = countS[i - 1] - 1;
                    countT[i] = countT[i - 1];
                }
                else countT[i] = countT[i - 1] + 1;
            }
        }
        
        PrintWriter out =  new PrintWriter(System.out);
        out.println(countS[countS.length - 1] + countT[countT.length - 1]);
        out.flush();
    }

    // 以下、ユーティリティ
    // n以上となる最小のインデックスを返す
    int binarySearch(int[] l, int n) {
        int left = -1;
        int right = l.length;
        while(right - left > 1) {
            int middle = left + (right - left) / 2;
            if(l[middle] >= n) right = middle;
            else left = middle;
        }

        return right;
    }

    Set<long[]> primeFactorize(long n){
        // たとえnが10^18でも約数の個数は400を超えないため素因数はもっと少なくなる
        Set<long[]> ret = new HashSet<>(500);
        for(long p = 2;p * p <= n;p++) {
            if(n % p != 0) continue;
            long num = 0;
            while(n % p == 0) {
                ++num;
                n /= p;
            }
            long[] arr = {p, num};
            ret.add(arr);
        }
        if(n != 1) {
            long[] arr = {n, 1};
            ret.add(arr);
        }

        return ret;
    }

    // nの約数を求める
    Set<Long> searchDivisor(long n){
        Set<Long> ret = new HashSet<>(200);
        for(long i = 1;i * i <= n;i++) {
            if(n % i == 0) {
                ret.add(Long.valueOf(i));
                if(i * i != n) ret.add(Long.valueOf(n / i));
            }
        }

        return ret;
    }

    // 素数判定
    public static boolean isPrime(long num) {
        if (num < 2) return false;
        else if (num == 2) return true;
        else if (num % 2 == 0) return false;

        double sqrtNum = Math.sqrt(num);
        for (int i = 3; i <= sqrtNum; i += 2) {
            if (num % i == 0) return false;
        }

        return true;
    }

    long nCm(long n, long m) {
        if(n < m) return 0l;
        long c = 1l;
        m = (n - m < m ? n - m : m);
        for(long ns = n - m + 1, ms = 1;ms <= m;ns ++, ms++) {
            c *= ns;
            c /= ms;
        }
        return c;
    }

    long modnCm(long n, long m, long mod) {
        if(n < m) return 0l;
        long upper = 1l;
        long downer = 1l;
        m = (n - m < m ? n - m : m);
        for(long ns = n - m + 1, ms = 1;ms <= m;ns ++, ms++) {
            upper = upper * ns % mod;
            downer = downer * ms % mod;
        }
        return upper * modInv(downer, mod) % mod;
    }

    // mod m での a の逆元 を計算する
    long modInv(long a, long m) {
        long b = m, u = 1, v = 0;
        while(b > 0) {
            long t = a / b;
            a -= t * b;
            long tmp = a;
            a = b;
            b = tmp;
            u -= t * v;
            tmp = u;
            u = v;
            v = tmp;
        }
        u %= m;
        if (u < 0) u += m;
        return u;
    }

    long modPow(long a, long n, long mod) {
        long res = 1l;
        while(n > 0l) {
            if(n % 2l == 1l) res = res * a % mod;
            a = a * a % mod;
            n /= 2l;
        }
        return res;
    }

}

class UnionFind {
    protected int[] parent;

    UnionFind(int n) {
        parent = new int[n];
        for(int i = 0;i < n;i++) parent[i] = i;
    }

    public int rootOf(int x) {
        if(parent[x] == x) return x;
        else return parent[x] = rootOf(parent[x]);
    }

    public boolean same(int x, int y) {
        return rootOf(x) == rootOf(y);
    }

    public void unite(int x, int y) {
        x = rootOf(x);
        y = rootOf(y);
        if(x == y) return;
        parent[x] = y;
        return;
    }
}

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            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;}
    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; 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() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDouble() { return Double.parseDouble(next());}
}