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: Accepted)

use std::vec::Vec;
use std::collections::VecDeque;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn read_touple() -> (usize, usize, char) {
  let v: Vec<String> = read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect();

  let mut res = (0, 0, 'a');
  res.0 = v[0].parse().ok().unwrap();
  if res.0 == 2 {
      res.1 = v[1].parse().ok().unwrap();
      res.2 = v[2].chars().collect::<Vec<char>>()[0];
  }
  res
}

fn main() {
  let s: Vec<char> = read::<String>().chars().collect();
  let n = s.len();
  let Q: usize = read();
  let mut querry = vec![];
  for i in 0..Q {
    let t = read_touple();
    querry.push(t);
  }

  let mut ans = VecDeque::new();

  for i in 0..n {
    ans.push_back(s[i]);
  }

  let mut rev = false;

  for &(t, f, c) in &querry {
    if t == 1 {
       rev = !rev;
       continue
    }

    // 初期状態から反転なし
    if rev == false {
       if f == 1 {
          ans.push_front(c);
       } else {
         ans.push_back(c);
       }
    }

    // 初期値から反転あり
    if rev == true {
      if f == 1 {
        ans.push_back(c);
      } else {
        ans.push_front(c);
      }
    }
  }

  let mut res = vec![];
  for e in ans {
    res.push(e);
  }

  let d: String = if rev == false {
    res.into_iter().collect()
  } else {
    res.into_iter().rev().collect()
  };


  //println!("{:?}",querry);
  println!("{}", d);
}