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 OCaml (4.02.3) (judge verdict: Accepted)

Scanf.scanf "%s %d %d" (fun s x y ->
    let len = String.length s in
    let rec loop i dir first cur xl yl =
        if i = len then (
            if cur > 0 then (
                if first < 0 then cur, xl, yl else
                if dir then first, (cur :: xl), yl
                       else first, xl, (cur :: yl)
            ) else first, xl, yl
        ) else (
            if s.[i] = 'F' then loop (i + 1) dir first (cur + 1) xl yl else
                let first, xl, yl =
                    if cur > 0 then (
                        if first < 0 then cur, xl, yl else
                        if dir then first, (cur :: xl), yl
                               else first, xl, (cur :: yl)
                    ) else max first cur, xl, yl
                in
                loop (i + 1) (not dir) first 0 xl yl
        )
    in
    let fx, xl, yl = loop 0 true (-1) 0 [] [] in
    let judge p l =
        let a = Array.make 8192 false in
        a.(0) <- true;
        List.iter (fun v ->
            for j = 8191 downto v do
                a.(j) <- a.(j) || a.(j - v)
            done
        ) l;
        let sum = List.fold_left (+) 0 l in
        let rec loop i =
            if i = 8192 then false else
                if a.(i) && abs (sum - i * 2) = abs p then true else loop (i + 1)
        in
        loop 0
    in
    print_endline @@ if judge (x - fx) xl && judge y yl then "Yes" else "No"
)