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: Time Limit Exceeded)

require 'prime'
MOD=10**9+7;  cnt=0; sum=0; prev=nil; can=true; h=Hash.new(0)
def gs() gets.chomp end
def gi() gets.chomp.to_i end
def gsmi() gets.chomp.split.map(&:to_i) end

def ar2(s1,s2,de=nil) Array.new(s1){Array.new(s2,de)} end
def ar3(s1,s2,s3,de=nil) Array.new(s1){Array.new(s2){Array.new(s3,de)}} end

def desc(ar) ar.sort!{|x,y| y<=>x} end
def min(a,b) a<=b ? a : b end
def max(a,b) a>=b ? a : b end
def sum(ar) ar.inject(:+) end
def ceil(a,b) (a.to_f/b).ceil end

def C(a,b) b==0||a==b ? 1 : (b=a-b if a/2<b;(a-b+1..a).inject(:*)/(1..b).inject(:*)) end
def rui(ar) s=[0]; ar.size.times{|i| s<<s[-1]+ar[i]}; s end
def rev_rui(ar) s=[0]; ar.size.times.reverse_each{|i| s<<s[-1]+ar[i]}; s end

def puts_yesno(b) puts(b ? 'Yes' : 'No') end
def putsend(s) puts s; exit end
def pu(h) begin; require './../nemi_lib/debug.rb'; puts_val(h); rescue LoadError; end; end
########### ( ˘ω˘ )ｽﾔｧ… ###########
s=gs
q=gi
fl=0

q.times do
  inp=gets.chomp.split
  if inp[0]=="1"
    fl=(fl+1)%2
  else
    f=inp[1].to_i
    c=inp[2]
    if f==1
      if fl==0
        s=c<<s
      else
        s=s<<c
      end

    else
      if fl==0
        s=s<<c
      else
        s=c<<s
      end
    end
  end
end

if fl==0
  puts s
else
  puts s.reverse
end