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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

int main(void) {
  string s;
  cin >> s;
  vector<char> S;
  for (int i = 0;i < s.size();i++) {
    S.push_back(s[i]);
  }
  
  
  int N;
  cin >> N;
  
  int rev_flag = 1;
  
  int cmd, sub_cmd;
  char w;
  for (int i = 0;i < N;i++) {
    cin >> cmd;
    if (cmd == 1) {
      rev_flag = (rev_flag == 1) ?  2 : 1;
    } else {
      cin >> sub_cmd >> w;
      if (sub_cmd + rev_flag == 3) S.push_back(w);
      else S.insert(S.begin(), w);
    }
  }
  
  if (rev_flag == 2) reverse(S.begin(), S.end());
  if (rev_flag == 2) {
    for (int i = S.size() - 1; i >= 0;i--) {
      cout << S[i];
    }
    cout << endl;
    return 0;
  }
  else {
  	for(int i = 0;i < S.size();i++) {
    	cout << S[i];
  	}
  	cout << endl;
  
  	return 0;
  }
  
  return -1;
}