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 Ruby (2.3.3) (judge verdict: Wrong Answer)

s = gets.chomp
q = gets.chomp.to_i

flag = 1

pre = ''
nex= ''

rev_cnt = 0

q.times do |i|
  query = gets.chomp.split

  if query.size == 3

    # 順方向かつ始めにつけるとき
    if  query[1] == 1 && flag == 1
      pre = pre + query[2]
    elsif query[1] == 0 && flag == 1
      # 順方向かつ後ろにつけるとき
      nex = nex + query[2]
    elsif query[1] == 1 && flag == 0
      nex= nex + query[2]
      # 反転してるとき、末尾につけるはpreにつけるのと一緒
    else
      pre = pre + query[2]
    end


  else
    rev_cnt += 1
    if flag == 0
      # flag==1を順方向とする
      flag = 1
    else
      # flag ==0は逆方向
      # 最後は反転する
      flag = 0
    end
  end


end

# if flag == 0
#   s = nex + s + pre
# end

if rev_cnt % 2 == 0
  pre.reverse!
else
  nex.reverse!
end

if flag == 1
  puts pre + s + nex
else
  puts nex + s.reverse + pre
end