Problem:
Score : 400 points

Problem Statement

Takahashi has a string S consisting of lowercase English letters.

Starting with this string, he will produce a new one in the procedure given as follows.

The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:

If T_i = 1: reverse the string S.

If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.

If F_i = 1 : Add C_i to the beginning of the string S.

If F_i = 2 : Add C_i to the end of the string S.

Help Takahashi by finding the final string that results from the procedure.

Constraints

1 \leq |S| \leq 10^5

S consists of lowercase English letters.

1 \leq Q \leq 2 \times 10^5

T_i = 1 or 2.

F_i = 1 or 2, if provided.

C_i is a lowercase English letter, if provided.

Input

Input is given from Standard Input in the following format:

S
Q
Query_1
:
Query_Q

In the 3-rd through the (Q+2)-th lines, Query_i is one of the following:

1

which means T_i = 1, and:

2 F_i C_i

which means T_i = 2.

Output

Print the resulting string.

Sample Input 1

a
4
2 1 p
1
2 2 c
1

Sample Output 1

cpa

There will be Q = 4 operations. Initially, S is a.

Operation 1: Add p at the beginning of S. S becomes pa.

Operation 2: Reverse S. S becomes ap.

Operation 3: Add c at the end of S. S becomes apc.

Operation 4: Reverse S. S becomes cpa.

Thus, the resulting string is cpa.

Sample Input 2

a
6
2 2 a
2 1 b
1
2 2 c
1
1

Sample Output 2

aabc

There will be Q = 6 operations. Initially, S is a.

Operation 1: S becomes aa.

Operation 2: S becomes baa.

Operation 3: S becomes aab.

Operation 4: S becomes aabc.

Operation 5: S becomes cbaa.

Operation 6: S becomes aabc.

Thus, the resulting string is aabc.

Sample Input 3

y
1
2 1 x

Sample Output 3

xy

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

import sys
#from collections import deque as dq
input = sys.stdin.readline

def gcd(a: int, b: int):
    """ https://cocodrips.hateblo.jp/entry/2014/03/05/143623
    """
    while b:
        a, b = b, a%b
    return a

def lcm(a: int, b: int):
    return a * b // gcd(a, b)


def main():
	#N,A,B = map(int, input().split())
	S = input().rstrip()
	Q = int(input())
	#vA = list(map(int, input().split()))
	#vX = [input().rstrip() for _ in [0,]*N]
	
	rev = 0		# reverse
	
	DF = "" #dq([]) # reversed Head!!
	DR = "" #dq([])
	for q in range(Q):
		tq = input().split()
		if tq[0]=="1":
			rev ^= 1
		else:
			T,F,C = tq
			if F=="1":
				# C to top
				if rev == 0:
					DF += C #DF.appendleft(C) 
				else:
					DR += C #DR.append(C)
			else:
				# C to bottom
				if rev == 0:
					DR += C #DR.append(C)
				else:
					DF += C #DF.appendleft(C)	
	
	
	if rev == False:
		res = DF[::-1] + S + DR
	else:
		res = DR[::-1] + S[::-1] + DF

	print(res)
	
if __name__ == "__main__":
    main()