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 Ruby (2.3.3) (judge verdict: Wrong Answer)

s = gets.chomp.split("")
x,y = gets.chomp.split(" ").map(&:to_i)
lx = [0]
ly = []
f = 1
s.each do |str|
  if str == "T"
    if f == 1
      ly << 0
    else
      lx << 0
    end
    f *= -1
  else
    if f == 1
      lx[-1] += 1
    else
      ly[-1] += 1
    end
  end
end
lx.delete(0)
ly.delete(0)
sx = lx.size
sy = ly.size
fx = false
fy = false
sumx = lx.inject(:+)
sumy = ly.inject(:+)
if sx > 0
  donex = Array.new(sx+1).map{Array.new(2*sumx+1)}
  donex[0][sumx] = true
  for i in 1..sx
    for j in 0..2*sumx
      if donex[i-1][j] == true
        donex[i][j+lx[i-1]] = true
        donex[i][j-lx[i-1]] = true
      end
    end
  end
  if donex[sx][sumx+x] == true
    fx = true
  end
elsif x == 0 && sx == 0
  fx = true
end
if sy > 0
  doney = Array.new(sy+1).map{Array.new(2*sumy+1)}
  doney[0][sumy] = true
  for i in 1..sy
    for j in 0..2*sumy
      if doney[i-1][j] == true
        doney[i][j+ly[i-1]] = true
        doney[i][j-ly[i-1]] = true
      end
    end
  end
  if doney[sy][sumy+y] = true
    fy = true
  end
elsif y == 0 && sy == 0
  fy = true
end

if fx && fy
  puts "Yes"
else
  puts "No"
end