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)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <utility>
#include <string>
#include <algorithm>
#include <map>
#include <vector>
#include <climits>
#include <numeric>
#include<stdio.h>
#include <queue>
#include <iomanip>
#include <float.h>
#include <set>
#include<functional>
#include <stack>
#include <time.h>
#include <climits>
#include <bitset>
using namespace std;
int  gcd(int a, int b) {
	if (a < b) gcd(b, a);
	int r;
	while ((r = a%b)) {
		a = b;
		b = r;
	}
	return b;
}
long long lcm(long long a, long long b) {
	return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long mod) {
	long long res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

long long modinv(long long a, long long mod) {
	long long b = mod, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t*b; swap(a, b);
		u -= t*v; swap(u, v);
	}
	u %= mod;
	if (u < 0) u += mod;
	return u;
}
namespace NTT {
	const int MOD = 998244353;  // to be set appropriately
	const long long PR = 3;     // to be set appropriately

	void trans(vector<long long> &v, bool inv = false) {
		int n = (int)v.size();
		for (int i = 0, j = 1; j < n - 1; j++) {
			for (int k = n >> 1; k >(i ^= k); k >>= 1);
			if (i > j) swap(v[i], v[j]);
		}
		for (int t = 2; t <= n; t <<= 1) {
			long long bw = modpow(PR, (MOD - 1) / t, MOD);
			if (inv) bw = modinv(bw, MOD);
			for (int i = 0; i < n; i += t) {
				long long w = 1;
				for (int j = 0; j < t / 2; ++j) {
					int j1 = i + j, j2 = i + j + t / 2;
					long long c1 = v[j1], c2 = v[j2] * w % MOD;
					v[j1] = c1 + c2;
					v[j2] = c1 - c2 + MOD;
					while (v[j1] >= MOD) v[j1] -= MOD;
					while (v[j2] >= MOD) v[j2] -= MOD;
					w = w * bw % MOD;
				}
			}
		}
		if (inv) {
			long long inv_n = modinv(n, MOD);
			for (int i = 0; i < n; ++i) v[i] = v[i] * inv_n % MOD;
		}
	}

	// C is A*B
	vector<long long> mult(vector<long long> A, vector<long long> B) {
		int size_a = 1; while (size_a < A.size()) size_a <<= 1;
		int size_b = 1; while (size_b < B.size()) size_b <<= 1;
		int size_fft = max(size_a, size_b) << 1;

		vector<long long> cA(size_fft, 0), cB(size_fft, 0), cC(size_fft, 0);
		for (int i = 0; i < A.size(); ++i) cA[i] = A[i];
		for (int i = 0; i < B.size(); ++i) cB[i] = B[i];

		trans(cA); trans(cB);
		for (int i = 0; i < size_fft; ++i) cC[i] = cA[i] * cB[i] % MOD;
		trans(cC, true);

		vector<long long> res((int)A.size() + (int)B.size() - 1);
		for (int i = 0; i < res.size(); ++i) res[i] = cC[i];
		return res;
	}
};

signed main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	string  s;
	cin >> s;
	int x, y;
	int nowx = 0, nowy = 0;
	cin >> x >> y;
	int now = 0;
	while (s[now] != 'T'&&now<s.length()) {
		now += 1;
		nowx += 1;
	}
	int muki = 0;
	int cnt[2] = { 0 };//x方向、y方向
	for (int i = now; i < s.length(); i++) {
		if (s[i] == 'T') {
			muki = 1 - muki;
		}
		else {
			cnt[muki] += 1;
		}
	}
	if (abs(x - nowx) <= cnt[0] && abs(y - nowy) <= cnt[1] && (abs(x - nowx) % 2 == cnt[0] % 2) && (abs(y - nowy) % 2 == cnt[1] % 2)) {
		cout << "Yes";
	}
	else {
		cout << "No";
	}
}