Problem:
Score : 100 points

Problem Statement

In AtCoder City, there are three stations numbered 1, 2, and 3.

Each of these stations is operated by one of the two railway companies, A and B. A string S of length 3 represents which company operates each station. If S_i is A, Company A operates Station i; if S_i is B, Company B operates Station i.

To improve the transportation condition, for each pair of a station operated by Company A and one operated by Company B, there will be a bus service connecting them.

Determine if there is a pair of stations that will be connected by a bus service.

Constraints

Each character of S is A or B.

|S| = 3

Input

Input is given from Standard Input in the following format:

S

Output

If there is a pair of stations that will be connected by a bus service, print Yes; otherwise, print No.

Sample Input 1

ABA

Sample Output 1

Yes

Company A operates Station 1 and 3, while Company B operates Station 2.

There will be a bus service between Station 1 and 2, and between Station 2 and 3, so print Yes.

Sample Input 2

BBA

Sample Output 2

Yes

Company B operates Station 1 and 2, while Company A operates Station 3.

There will be a bus service between Station 1 and 3, and between Station 2 and 3, so print Yes.

Sample Input 3

BBB

Sample Output 3

No

Company B operates all the stations. Thus, there will be no bus service, so print No.

Solution in PyPy3 (2.4.0) (judge verdict: Runtime Error)

S = input()
Q = int(input())
head = True # True: left side is head
for i in range(Q):
  Query_i = list(map(str, input().split()))
  if Query_i[0]=='1':
    head = not head
  elif Query_i[1] == '1' and head:
    S = Query_i[2] + S
  elif Query_i[1] == '1' and not head:
    S = S + Query_i[2]
  elif Query_i[1] == '2' and head:
    S = S + Query_i[2]
  else Query_i[1] == '2' and not head:
    S = Query_i[2] + S
  if not head:
    S = S[::-1]
print(S)