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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program String_Formation
    implicit none
    character S*400005,C*1,query*100
    character ans(-400005:400005)
    integer Q,l,r,i,T,rev,F
    integer sf(100)

    read *,S,Q
    l = 0
    r = len_trim(S)+1
    rev = 0
    ans(1:) = transfer(S,ans,size=len(S))

    do i = 1, Q
        read (*,'(a)')query
        if(query(1:1) == '1') then
             rev = rev+1
        else
            read (query,*)T,F,C 
            if(F == 1) then
                if(mod(rev,2) == 1) then
                    ans(r:r) = C
                    r = r+1
                else
                    ans(l:l) = C
                    l = l-1
                endif
            else
                if(mod(rev,2) == 0) then
                    ans(r:r) = C
                    r = r+1
                else
                    ans(l:l) = C
                    l = l-1
                endif
            endif
        endif
    enddo

    if(mod(rev,2) == 0) then
        print *,ans(l+1:r-1)
    else
        print *,ans(r-1:l+1:-1)
    endif


end program String_Formation