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::collections::VecDeque;
use std::io::stdin;
use std::str::FromStr;

fn read_line<T>() -> Vec<T>
where
    T: FromStr,
    <T as std::str::FromStr>::Err: std::fmt::Debug,
{
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    s.trim()
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect()
}

fn read_str_as_char_vec() -> VecDeque<char> {
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    s.trim().chars().collect()
}

fn read_str_as_T<T>() -> T
where
    T: FromStr,
    <T as std::str::FromStr>::Err: std::fmt::Debug,
{
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    s.trim().parse().unwrap()
}

fn main() {
    let mut s = read_str_as_char_vec();
    let q: usize = read_str_as_T();

    let mut rev = 1;
    for _i in 0..q {
        // let in_vec: Vec<char> = read_line();
        let mut temp = String::new();
        stdin().read_line(&mut temp).ok();
        let mut in_iter = temp.trim().split_whitespace();
        let mut t = in_iter.next().unwrap().chars();

        if t.next().unwrap() == '1' {
            rev *= -1;
        } else {
            let f: usize = in_iter.next().unwrap().parse().unwrap();
            let mut p = in_iter.next().unwrap().chars();
            if f == 1 {
                if rev == 1 {
                    s.push_front(p.next().unwrap());
                } else {
                    s.push_back(p.next().unwrap());
                }
            } else {
                if rev == -1 {
                    s.push_front(p.next().unwrap());
                } else {
                    s.push_back(p.next().unwrap());
                }
            }
        }
    }

    let mut res: String = String::new();
    if rev == 1 {
        for &c in s.iter() {
            res.push(c);
        }
    } else {
        for &c in s.iter().rev() {
            res.push(c);
        }
    }

    println!("{}", res);
}