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)

fn main() {
    let s: String = read();
    let n: usize = read();
    let mut rev = false;
    let mut v: Vec<Vec<char>> = vec![vec![]; 2]; // index: 0 のときは front, index: 1のときは back
    for _ in 0..n {
        let op: usize = read();
        match op {
            1 => rev = !rev,
            2 => {
                let f: usize = read();
                let c = read::<String>().chars().next().unwrap();
                match f {
                    // reverse 状態のときは、 front end も行動を反転する
                    1 => v[if rev { 1 } else { 0 }].push(c),
                    2 => v[if rev { 0 } else { 1 }].push(c),
                    _ => unreachable!(),
                }
            }
            _ => unreachable!(),
        }
    }
    let c: String = if rev { s.chars().rev().collect() } else { s };
    let f: String = v[if rev { 1 } else { 0 }].iter().cloned().rev().collect();
    let b: String = v[if rev { 0 } else { 1 }].iter().cloned().collect();
    println!("{}{}{}", f, c, b);
}

use std::io::stdin;
use std::io::Read;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let cin = stdin();
    let cin = cin.lock();
    let token: String = cin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}