Problem:
Score : 500 points

Problem Statement

A robot is put at the origin in a two-dimensional plane.
Initially, the robot is facing in the positive x-axis direction.

This robot will be given an instruction sequence s.
s consists of the following two kinds of letters, and will be executed in order from front to back.

F : Move in the current direction by distance 1.

T : Turn 90 degrees, either clockwise or counterclockwise.

The objective of the robot is to be at coordinates (x, y) after all the instructions are executed.
Determine whether this objective is achievable.

Constraints

s consists of F and T.

1 \leq |s| \leq 8 000

x and y are integers.

|x|, |y| \leq |s|

Input

Input is given from Standard Input in the following format:

s
x y

Output

If the objective is achievable, print Yes; if it is not, print No.

Sample Input 1

FTFFTFFF
4 2

Sample Output 1

Yes

The objective can be achieved by, for example, turning counterclockwise in the first T and turning clockwise in the second T.

Sample Input 2

FTFFTFFF
-2 -2

Sample Output 2

Yes

The objective can be achieved by, for example, turning clockwise in the first T and turning clockwise in the second T.

Sample Input 3

FF
1 0

Sample Output 3

No

Sample Input 4

TF
1 0

Sample Output 4

No

Sample Input 5

FFTTFF
0 0

Sample Output 5

Yes

The objective can be achieved by, for example, turning counterclockwise in the first T and turning counterclockwise in the second T.

Sample Input 6

TTTT
1 0

Sample Output 6

No

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

///Bismillahir Rahmanir Rahim
#include<bits/stdc++.h>

#define                           ll                          long long
#define                           int                         ll
#define                           fi                          first
#define                           si                          second
#define                           mp                          make_pair
#define                           pb                          push_back
#define                           pi                          pair<ll,ll>
#define                           clr(x)                      memset(x,0,sizeof(x));
#define                           f(i,l,r)                    for(int i=l;i<=r;i++)
#define                           rf(i,r,l)                   for(int i=r;i>=l;i--)
#define                           done(i)                     cout<<"done = "<<i<<endl;
#define                           show(x,y)                   cout<<x<<" : ";for(auto z:y)cout<<z<<" ";cout<<endl;
#define                           fast                        ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

using namespace std;

const ll inf=1e18;
const int mod=1e9+7;
const int M=100005;

 main()

{
    fast
    string s;
    cin>>s;
    int x,y;
    cin>>x>>y;
    int nxt;
    int k=0,sum1=0,sum2=0;
    int sz=s.size();
    f(i,0,sz-1)
    {
        if(s[i]=='T')
        {
            nxt=i+1;
            break;
        }
        else k++;
    }
    int cnt=0,turn=1;
    s=s+"T";
    f(j,nxt,sz)
    {
        if(s[j]=='T')
        {

            if(turn==1)sum2+=cnt;
            else sum1+=cnt;
            cnt=0;
            turn=1-turn;
        }
        else
        {
            cnt++;
        }
    }
    //cout<<k<<" "<<sum1<<" "<<sum2<<endl;
    if((x-(k)-sum1)%2==0 && (x-k-sum1)>=0 && (x-k-sum1)<=(sum1+k))
    {
        if((y-sum2)%2==0 && (y-sum2)>=0 && (y-sum2)<=sum2)
        {
            cout<<"Yes"<<endl;
            return 0;
        }
    }
    cout<<"No"<<endl;
    return 0;

}