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: Runtime Error)

#include<bits/stdc++.h>
#include<algorithm>
#include<math.h>
#include<string>
#include<iostream>
#include<vector>
#include<cstdlib>
#include<stdio.h>
#include<cmath>
#include<limits>
#include<iomanip>
#include<cstdlib> 
#include<ctime>
#include<random>


using namespace std;

template <class T> using V = vector<T>;

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

const long long INF = 1LL << 60;

using ll = long long;
using db = long double;
using st = string;
using ch = char;
using bl = bool;
using vll = V<ll>;
using vpll =V<pair<ll,ll>>;
using vst = V<st>;
using vdb = V<db>;
using vch = V<ch>;
using vbl = V<bl>;
using graph = V<V<int>>;
using pq = priority_queue<ll>;



#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define rFOR(i,a,b) for(ll i=(a);i>=(b);i--)
#define oFOR(i,a,b) for(ll i=(a);i<(b);i+=2)
#define bgn begin()
#define en end()
#define SORT(a) sort((a).bgn,(a).en)
#define REV(a) reverse((a).bgn,(a).en)
#define M(a,b) max(a,b)
#define rM(a,b) min(a,b)
#define fi first
#define se second
#define sz size()
#define gcd(a,b) __gcd(a,b)
#define co(a) cout<<a<<endl;
#define pb(a) push_back(a);
#define ALL(a)  (a).begin(),(a).end()




ll Sum(ll n) {
ll m=0;
  while(n){
    m+=n%10;
    n/=10;
  }
  return m;
}



const int MAX = 510000;
const int MOD = 1000000007;
 
long long fac[MAX], finv[MAX], inv[MAX];
 
void Comuse() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}


#define comuse Comuse()

ll combi(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll perm(int n,int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] % MOD) % MOD;
}

ll modpow(ll a,ll n,ll mod){
  ll ans=1;
  while(n>0){
    if(n&1){
      ans=ans*a%mod;
    }

    a=a*a%mod;
    
    n>>=1;
  }

  return ans;
}

ll modinv(ll a, ll mod) {
    return modpow(a, mod - 2, mod);
}

ll modcombi(int n,int k,int mod){
  ll ans=1;
  for(ll i=n;i>n-k;i--){
    ans*=i;
    ans%=mod;
  }
  for(ll i=1;i<=k;i++){
    ans*=modinv(i,mod);
    ans%=mod;
  }

  return ans;
}

ll lcm(ll a,ll b){
  ll n;
  n=a/gcd(a,b)*b;
  return n;
}


vll div(ll n){
  vll ret;
  for(ll i=1;i*i<=n;i++){
    if(n%i==0){
      ret.push_back(i);
      if(i*i!=n){
        ret.push_back(n/i);
      }
    }
  }
  SORT(ret);
  return (ret);
}

vector<bool> isprime(MAX+100,true);

void primeuse(){
  isprime[0]=false;
  isprime[1]=false;
  for(int i=2;i<MAX+50;i++){
    int up=sqrt(i)+1;
    for(int j=2;j<up;j++){
      if(i%j==0){
        isprime[i]=false;
      }
    }
  }
}



void Solve();


const int MAX_N = 131072;
//segment tree 
int NN;
int seg[MAX_N*2-1];
void seguse(){
  for(int i=0;i<2*NN-1;i++){
    seg[i]=INT_MAX;
  }
}


signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<setprecision(20)<<fixed;
  Solve();
}



/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/
//segtreeの葉の先頭の添え字はN-1
void Solve(){
  //ll N=262144;
  //vll segtree(2*N-1);
  st x;
    cin>>x;
  ll n=x.sz;
  ll count=1;
  vll A(0);
  FOR(i,1,n){
    if(x[i-1]!=x[i]){
      A.push_back(count);
      count=1;
    }
    else{
      count++;
    }
  }
  A.push_back(count);
  ll how=0;
  if(x[0]=='T'){
    how++;
  }

  ll ans=n;
  ll s=0;
  ll t=0;
  FOR(i,how,n){
    s+=A[i];
    if(s<A[i+1]){
      ans-=2*s;
      s=0;
    }
    else{
      ans-=2*A[i+1];
      s-=A[i+1];
    }
    i++;
  }
  cout<<ans<<"\n";
}