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)

# 高橋君は、英小文字から成る文字列 S を持っています。
# この S から始めて、ある与えられた手順に従って文字列を作ることにしました。
# 手順は Q 回の操作から成ります。操作 i(1≤i≤Q) では、まず整数 Ti が与えられます。
# Ti=1 のとき : 文字列 S の前後を反転する。
# Ti=2 のとき : 追加で整数 Fi と英小文字 Ci が与えられる。
# Fi=1 のとき : 文字列 S の先頭に Ci を追加する。
# Fi=2 のとき : 文字列 S の末尾に Ci を追加する。
# 高橋君のために、手順の後に最終的にできる文字列を求めてあげてください。

s = gets.chomp
t = String.new
q = gets.to_i

is_reverse = false
q.times do
    input = gets.chomp.split
    case input[0].to_i
    when 1
        is_reverse = !is_reverse
    when 2
        case input[1].to_i
        when 1
            case is_reverse
            when false
                t << input[2]
            when true
                s << input[2]
            end
        when 2
            case is_reverse
            when false
                s << input[2]
            when true
                t << input[2]
            end
        end
    end
end

t.reverse!
str = t + s
puts str