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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

S = input()
Q = int(input())

def add(c ,head, tail, r, ri, f):
    r_m = ri % 2
    # print(r, ri)
    if (r-ri) % 2 == 1:
        f = 1 if f == 2 else 2
    # print(c, r_m, f)
    # if (r_m == 0 and f == 1) or (r_m == 1 and f == 2):
    if f==1:
        head = [c] + head
    else:
        tail = tail + [c]

    return head, tail

def put(c, f, ri, m):
    m.append((c, f, ri))
    return m

def join(s, head, tail, r):
    s_l = list(s)
    if r % 2 == 1:
        s_l = s_l[::-1]
    return ''.join(head + s_l + tail)

# def reverse(head, tail):
#     return tail[::-1], head[::-1]


r = 0
head = []
tail = []
m = []
for _ in range(Q):
    q = [x for x in input().split(' ')] 
    if len(q) == 1:
        r += 1
    else:
        _, f, c = q
        f = int(f)
        m = put(c, f, r, m)

for c, f, ri in m:
    head, tail = add(c, head, tail, r, ri, f)

ans = join(S, head, tail, r)
print(ans)