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)

def s()
    gets().chomp
  end
  def i()
    gets.to_i
  end
  def li()
    gets.split.map(&:to_i)
  end
  def ls()
    gets.split
  end
  def ili()
    n = gets.to_i
    as = gets.split.map(&:to_i)
    [n, as]
  end
  def inli()
    n = gets.to_i
    as = []
    n.times {
      as << gets.split.map(&:to_i)
    }
    [n, as]
  end
  class Array
    def sum
      total = 0
      self.each do |item|
        total += item
      end
      total
    end
  end
  
  def eratosthenes(size)
    prime_list = []
    is_prime = Array.new(size, true)
    is_prime[0] = false
    is_prime[1] = false
  
    2.upto(size-1){|i|
      if is_prime[i]
        prime_list << i
        (2).upto((size.to_f/i).ceil-1){|j|
          is_prime[j*i] = false
        }
      end
    }
    [prime_list, is_prime]
  end
  
  def dijkstra(s, e, cost)
    size = cost.length
    if s==e
      return 0
    end
    dp = {}
    dp[s] = 0
    undefined_node = {}
    size.times{|i|
      if i != s
        undefined_node[i] = cost[s][i]
      end
    }
    while undefined_node.length != 0
      min = undefined_node.min_by{|_, v| v}
      stock = {}
      temp = undefined_node.clone
      temp.each{|k, v|
        if min[1] == v
          undefined_node.delete(k)
          if k == e
            return v
          end
          dp[k] = v
          stock[k] = v
        end
      }
  
      stock.each{|k, v|
        undefined_node.each{|k2, v2|
          undefined_node[k2] = v + cost[k][k2] < v2 ?  v + cost[k][k2] : v2
        }
      }
    end
  end
  
  def get_bits(len)
    res =  []
    (2**len).times{|i|
      res << i.to_s(2).rjust(len, '0')
    }
    res
  end
  
  def get_GCD(m, n)
    tmp = [n, m].max
    n = [n, m].min
    m = tmp
    while n != 0
      tmp = m % n
      m = n
      n = tmp
    end
    m
  end
  
  def matrix(n, m=n, init=0)
    Array.new(n).map{Array.new(m,init)}
  end


S = s()
n = i()
qs = []
n.times{
    qs << ls()
} 
is_reversed = false
tail = ""
head = ""
qs.each{ |q|
  if q[0] == "1" 
    is_reversed = !is_reversed
  else
    if is_reversed 
        if q[1] == "1" 
            
            tail << q[2]
        else
            head << q[2]
        end
    else
        if q[1] == "1" 
            head << q[2]
        else
            tail << q[2]
        end
    end
  end
}
if is_reversed
    puts tail.reverse + S + head
else
    puts head.reverse + S + tail
end