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: Accepted)

#include <vector>
#include <iostream>
#include <algorithm>
#include <string>
#include <sstream>
#include <functional>
#include <unordered_map>
#include <set>

#include <cassert>
using namespace std;

int main(void){

	string s;
	cin >> s;
	vector<vector<short>> n(s.size()+2);

	int dx,dy;
	cin >> dx >> dy;

	int ofx=0;
	int i;
	for(i=2;i<n.size();++i){
		int si_now = i-2;
		if(s[si_now]=='T'){
			dx = abs(dx);
			dy = abs(dy);
			i+=1;
			break;
		}else{
			dx-=1;
		}
	}
	
	int d = 1; //0=x,1=y,最初は1回方向転換済みなのでy
	int ix=1,iy=0;
	n[0].push_back(0); //y
	n[1].push_back(0); //x
	for(;i<n.size();++i){
		int si_now = i-2;
		if (s[si_now]=='T'){
			d = 1 - d;
		}else{ // 'F'
			auto id = (d==0)?ix:iy;
			auto dd = (d==0)?dx:dy;
			if(s[si_now-1]=='T'){
				set<short> nnext;
				for(int k=0;k<n[id].size();++k){
					const auto &v = n[id][k];
					auto in1=v+1,in2=v-1;
					nnext.insert(in1);
					nnext.insert(in2*-1);
				}
				n[i].reserve(nnext.size());
				for(auto it=nnext.begin();it!=nnext.end();++it){ n[i].push_back(*it); }
			}else{
				n[i].resize(n[i-1].size());
				for(int k=0;k<n[i].size();++k){
					const auto &v = n[id][k];
					n[i][k] = v+1;
				}
				sort(n[i].begin(),n[i].end());
				n[i].erase( unique(n[i].begin(), n[i].end()), n[i].end());
			}
			if(d==0){ ix=i; }else{ iy=i; }
		}
	}

	if(binary_search(n[ix].begin(),n[ix].end(),dx) && binary_search(n[iy].begin(),n[iy].end(),dy)){
		std::cout << "Yes" << endl;
	}else{
		std::cout << "No" << endl;
	}

	return 0;
}