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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define intll int long long
const int INF = 1e9;
const int MOD = 1e9 + 7;

int main() {
  string s ;
  deque <char>  data ;
  int n ;
  cin >> s >> n ;
  rep(i,s.size()) data.push_back(s.at(i)) ;
  int now = 0 ;
  
  rep(i,n){
    int T ;
    cin >> T ;
    if(T == 1) now ++ ;
    else{
      int F ;
      char c ;
      cin >> F >> c ;
      if(F == 1){ 
        if(now%2 == 0) data.push_back (c) ;
        else           data.push_front(c) ;
      }
      else {
        if(now%2 != 1) data.push_back (c) ;
        else           data.push_front(c) ;
    }
  }
    
}
 
  if(now % 2 == 1)   rep(i,data.size()) cout << data.at(i) ;
  else               rep(i,data.size()) cout << data.at(data.size() - i-1) ;
  
}