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 PyPy3 (2.4.0) (judge verdict: Runtime Error)

# https://atcoder.jp/contests/abc082/tasks/arc087_b


def run_length_encoding(s):
    '''
    連長圧縮を行う
    s ... iterable object e.g. list, str
    return
    ----------
    s_composed,s_num,s_idx
    それぞれ、圧縮後の文字列、その文字数、その文字が始まるidx
    '''
    s_composed = []
    s_sum = []
    s_idx = [0]
    pre = s[0]
    cnt = 1
    for i, ss in enumerate(s[1:], start=1):
        if pre == ss:
            cnt += 1
        else:
            s_sum.append(cnt)
            s_composed.append(pre)
            s_idx.append(i)
            cnt = 1
            pre = ss
    s_sum.append(cnt)
    s_composed.append(pre)
    # assert len(s_sum) == len(s_composed)
    return s_composed, s_sum, s_idx

# データに関しては事前にランレングス圧縮しておくと楽かもね


s = input()
x, y = map(int, input().split())
s_composed, s_sum, _ = run_length_encoding(s)

if s_composed[0] == 'T':  # 必ずTから始まるようにしたい
    s_composed.insert(0, 'F')
    s_sum.insert(0, 0)
# 前処理 xの移動履歴とyの移動履歴を作る
is_x = True
hist_x = []
hist_y = []
for s, n in zip(s_composed, s_sum):
    if s == 'T':
        if n & 1:  # xyの状態を反転させる
            is_x = not is_x
    else:
        if is_x:
            hist_x.append(n)
        else:
            hist_y.append(n)

'''
dpx[i][j]...x座標がjであることが可能か？ x方向の移動についてi回目の移動のあとに。
dpy[i][j]...y座標がjであることが可能か？ y方向の移動についてi回目の移動のあとに。

更新則
xについての移動,距離d移動するとすると
dpx[i+1][j+d]=dpx[i+1][j+d] or dpx[i][j]
dpx[i+1][j-d]=dpx[i+1][j-d] or dpx[i][j]
(配るDP)で更新すると楽かな？
yも同様

初期条件
dpx[:][:]=False
dpx[0][d_x0]=True #xの0回目の移動。

dpy[:][:] = False
dpy[0][d_y0]=True 簡単のために
dpy[0][-d_y0]=True
'''

if hist_x:
    x_range = 2 * (sum(hist_x) + 100 + max(hist_x))  # ここが0になるときがある
    shift = x_range // 2
    dpx = [[False] * x_range for _ in range(len(hist_x))]
    dpx[0][hist_x[0] + shift] = True
    for i in range(1, len(hist_x)):
        d = hist_x[i]
        for j in range(x_range):
            if j + d < x_range:
                dpx[i][j] = dpx[i][j] or dpx[i - 1][j + d]
            if -1 < j - d:
                dpx[i][j] = dpx[i][j] or dpx[i - 1][j - d]
    ansx = dpx[-1][x + shift]
# else:
#     ansx = x == 0

if hist_y:
    y_range = 2 * (sum(hist_y) + 100 + max(hist_y))  # ここが0になるときがある
    shift = y_range // 2
    dpy = [[False] * y_range for _ in range(len(hist_y))]
    dpy[0][hist_y[0] + shift] = True
    dpy[0][-hist_y[0] + shift] = True
    for i in range(1, len(hist_y)):
        d = hist_y[i]
        for j in range(y_range):
            if j + d < y_range:
                dpy[i][j] = dpy[i][j] or dpy[i - 1][j + d]
            if -1 < j - d:
                dpy[i][j] = dpy[i][j] or dpy[i - 1][j - d]
    ansy = dpy[-1][y + shift]
else:
    ansy = y == 0

print('Yes' if ansx and ansy else 'No')


# print(ansx, ansy)
# print(*dpy, sep='\n')