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::str::FromStr;

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

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

fn read_vec2<T: FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

pub struct Foo {
    ch: char
}

pub struct ParseCharError {
    kind: CharErrorKind,
}

impl ParseCharError {
    pub fn __description(&self) -> &str {
        match self.kind {
            CharErrorKind::EmptyString => "cannot parse char from empty string",
            CharErrorKind::TooManyChars => "too many characters in string",
        }
    }
}

enum CharErrorKind {
    EmptyString,
    TooManyChars,
}

impl FromStr for Foo {
    type Err = ParseCharError;

    fn from_str(s: &str) -> Result<Self, Self::Err> {
        let mut chars = s.chars();
        match (chars.next(), chars.next()) {
            (None, _) => Err(ParseCharError { kind: CharErrorKind::EmptyString }),
            (Some(c), None) => Ok(Foo { ch: c }),
            _ => Err(ParseCharError { kind: CharErrorKind::TooManyChars }),
        }
    }
}

fn main() {
    let s: String = read();
    let n: u32 = read();
    let q: Vec<Vec<Foo>> = read_vec2(n);
    let mut ans = s.chars().collect::<Vec<char>>();
    let mut is_reverse = false;
    for query in q {
        if query[0].ch == '1' {
            is_reverse = if is_reverse { false } else { true };
        } else {
            if query[1].ch == '1' {
                if is_reverse {
                    ans.push(query[2].ch);
                } else {
                    ans.insert(0, query[2].ch);
                }
            } else {
                if is_reverse {
                    ans.insert(0, query[2].ch);
                } else {
                    ans.push(query[2].ch);
                }
            }
        }
    }
    if is_reverse {ans.reverse()}
    println!("{}", ans.into_iter().collect::<String>());
}