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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

use std::io::{self, BufRead};

fn main() {
  let stdin = io::stdin();
  let mut i = 0;
  let mut s = String::new();
  let mut size: i32 = -1;
  for line in stdin.lock().lines() {
    match i {
      0 => s = line.unwrap(),
      1 => size = line.unwrap().parse().unwrap(),
      _ => {
        let xs = line.unwrap();
        let ys: Vec<&str> = xs.split(' ').collect();
        match ys[0] {
          "1" => s = s.chars().rev().collect::<String>(),
          "2" => match ys[1] {
            "1" => {
              let mut tmp = String::from(ys[2]);
              tmp.push_str(&s);
              s = tmp;
            },
            "2" => s.push_str(ys[2]),
            _ => (),
          },
          _ => (),
        }
      },
    }
    i += 1;
    if size >= 0 && size == i - 2 {
      break;
    }
  }
  println!("{}", s);
}