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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define rep(i,n) for(long long int (i)=0;(i)<(int)(n);(i)++)
#define rrep(i,a,b) for(long long int i=(a);i<(b);i++)
#define rrrep(i,a,b) for(long long int i=(a);i>=(b);i--)
#define all(v) (v).begin(), (v).end()
#define pb(q) push_back(q)
#define P pair<int,int>
#define Abs(a,b) max(a,b)-min(a,b)
#define YES(condition) if(condition){cout << "YES" << endl;}else{cout << "NO" << endl;}
#define Yes(condition) if(condition){cout << "Yes" << endl;}else{cout << "No" << endl;}
#define Cout(x) cout<<(x)<<endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) << endl
#define possible(n) cout << ((n) ? "possible" : "impossible"  ) << endl
#define Size(n) (n).size()

typedef long long ll;

using namespace std;

const int INF = 1e9,MOD = 1e9 + 7;
const ll LINF = 1e18;

//long long int kaizyo(long long int hh){
  //  cmp=1;
    //while(hh>1){
      //  cmp=(cmp*hh)%MOD;
        //hh--;
    //}
    //return cmp;
//}
//long long int ruizyo(long long int aa, long long int bb){
  //  if(aa==0){
    //    return 1;
    //}
    //else if(aa%2==0){
      //  long long int tt=ruizyo(aa/2,bb);
        //return (tt*tt)%MOD;
    //}
    //else{
      //  return (ruizyo(aa-1,bb)*bb)%MOD;
    //}
//}フェルマ－のア


//while(x!=0){
            //sum+=x%10;
           // x/=10;
        //}
        //各桁の和

//pair<int,int> p[100000];
//P r[100000];
//cin >> tmp;
//p[i]=make_pair(tmp,i);
//cout << p[i].second+1 << endl;//ペアの右側つまりiを出力

//s.find(w[i])==string::npos
//findの使い方

//for(int i=0;i<n;i++){
   // b[i]=x%2;
    //x/=2;
  //}二進数

//visited=vector<vector<bool>>(10,vector<bool>(10,false));アを全部falseに


long long int n,cnt=0,ans=1,a[10000000],b[10000000],c,d,cmp,cmpp,m,h,w,x,y,sum=0,pos;
int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
string alph("abcdefghijklmnopqrstuvwxyz"),s;
bool Sfl=false;

int main(void){
       cin.tie(0);
    ios::sync_with_stdio(false);

        cin>>s;
        ans=Size(s);
        rep(i,Size(s)){
                if(s[i]=='S'){
                        cnt++;
                        Sfl=true;
                }
                else if(Sfl){
                        cnt--;
                        ans-=2;
                }
        }
        Cout(ans);
return 0;
}