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)

lines = readlines.map{|line| line.split(' ')}
a=lines[0][0]
b=lines[1][0].to_i
c=[]
d=0
e=0
for i in 2..b+1 do
  if lines[i][0].to_i != 1 then
    d=1
  else
    d=0
  end
  if lines[i][0].to_i == 1 && d==0 then
    e=e+1
  else
    if c.length>0 && i>2 && lines[i-1][0].to_i==1 then
      for j in c do
        if (j[1].to_i+e)%2==1 then
          a=j[2] << a
        else
          a=a << j[2]
        end
      end
      e=0
      c=[]
      c.push(lines[i])
    else
      c.push(lines[i])
    end
  end
end
if d != 1 then
  for j in c do
    if (j[1].to_i+e)%2==1 then
      a=j[2] << a
    else
      a=a << j[2]
    end
  end
end
puts a