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: Accepted)

/*
	DATE:		2020-03-10 11:19:28
	NAME:		
	PROBLEM:	ABC158D
*/
#include <bits/stdc++.h>

using namespace std;

const int MAXN = 1e5;

int a[MAXN + 7];

char s0[MAXN + 7];

int main() {
  scanf("%s", s0 + 1);
  
  deque<char> dq;
  int len = strlen(s0 + 1);
  for (int i = 1; i <= len; i++) {
    dq.push_back(s0[i]);
  }

  int q;
  scanf("%d", &q);

  bool is_reversed = false;

  while (q--) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      is_reversed ^= 1;
    }
    else {
      int f;
      char ch;
      scanf("%d %c", &f, &ch);
      f--;
      f ^= is_reversed;
      if (f == 0) {
        dq.push_front(ch);
      }
      else {
        dq.push_back(ch);
      }
    }
  }

  if (is_reversed) {
    reverse(dq.begin(), dq.end());
  }

  for (int i = 0; i < dq.size(); i++) {
    printf("%c", dq[i]);
  }
  printf("\n");

  return 0;
}