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: Wrong Answer)

use std::io::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .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")
}

pub fn read_n<T: FromStr>(n: usize) -> Vec<T> {
    read_n_logic::<T>(n, vec![])
}

pub fn read_n_logic<T: FromStr>(n: usize, mut a: Vec<T>) -> Vec<T> {
    match n {
        0 => a,
        _ => {
            a.push(read::<T>());
            read_n_logic(n - 1, a)
        }
    }
}
use std::cmp::{max, min};
use std::collections::VecDeque;

fn main() {
    let mut vd: VecDeque<char> = VecDeque::new();
    let mut isReverse = false;
    let c: String = read();
    let cc: Vec<char> = c.chars().collect();
    vd.push_back(cc[0]);

    let n: i64 = read();
    for i in 0..n {
        let f: usize = read();
        if f == 1 {
            isReverse = !isReverse;
        } else {
            let w: usize = read();
            let ch: String = read();
            let cc: Vec<char> = ch.chars().collect();
            let isFront = if !isReverse { w == 1 } else { w == 2 };
            if isFront {
                vd.push_front(cc[0])
            } else {
                vd.push_back(cc[0])
            }
        }
    }

    let mut ans = "".to_string();
    while !vd.is_empty() {
        if !isReverse {
            ans += &vd.pop_front().unwrap().to_string()
        } else {
            ans += &vd.pop_back().unwrap().to_string()
        }
    }
    println!("{}", ans);
}