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)

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rep1(i,n) for(int i=1;i<=(int)(n);++i)
#define rep11(i,n) for(int i=1;i<(int)(n);++i)
#define repo(i,o,n) for(int i=o;i<(int)(n);++i)
#define repm(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define sperase(v,n) (v).erase(remove(all(v), n), (v).end())
#define vdelete(v) (v).erase(unique(all(v)), (v).end())
#define pb(n) push_back(n)
#define mp make_pair
#define MOD 1000000007
#define INF LONG_LONG_MAX

int X,Y,now,cnt;
bool memox[16001], memoy[16001];
string s;
vector<int> x,y;

void kr(vector<int>& v, bool memo[16001]) {
    memo[8000] = true;
    bool f = false;
    for(auto i : v) {
        bool tmp[16001]={};
        rep(j,16000) {
            if (memo[j]) {
                if (f && j-i >= 0) tmp[j-i] = true;
                if (j+i <= 16000) tmp[j+i] = true;
            }
        }
        rep(j,16001) memo[j] |= tmp[j];
        f = true;
    }
}

signed main() {
    cin >> s >> X >> Y;
    for(auto i : s) {
        if (i == 'F') cnt++;
        else {
            if (now%2) y.pb(cnt);
            else x.pb(cnt);
            cnt = 0;
            now++;
        }
    }
    if (cnt) {
        if (now%2) y.pb(cnt);
        else x.pb(cnt);
    }
    kr(x, memox);
    kr(y, memoy);
    if (memox[8000+X] && memoy[8000+Y]) cout << "Yes" << endl;
    else cout << "No" << endl;
}