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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <algorithm>

using namespace std;

//1なら先頭，0なら末尾から見ていく
int top = 1;

//文字列sの先頭か末尾に文字列を追加する
string push_string(string s, char q[3]){
  if(top == 1){
    if(q[1] == '1')
      s = q[2]+s;
    else
      s = s+q[2];
  }
  else{
    if(q[1] == '1')
      s = s+q[2];
    else
      s = q[2]+s;
  }
  return s;
}


int main(void){
  string S;   cin >> S;
  int Q;   cin >> Q;
  char query[Q][3];
  int i, j;
  for(i=0; i<Q; i++){
    cin >> query[i][0];
    if(query[i][0] == '2')
      cin >> query[i][1] >> query[i][2];
  }

  //各クエリの処理
  for(i=0; i<Q; i++){
    //反転させるとき
    if(query[i][0] == '1'){
      top = !top;
    }
    //追加する時とき
    else
      S = push_string(S, query[i]);
  }

  if(top == 1)
    cout << S << endl;
  else{
    for(i=S.length()-1; i>=0; i--)
      cout << S[i];
    cout << endl;
  }
  
  return 0;
}