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)

#![allow(unused_imports)]pub mod pcl{pub mod stdin{use std::io::prelude::*;use std::io::stdin;use std::str::FromStr;pub fn read_line()->String{let mut s=String::new();stdin().read_line(&mut s).unwrap();s.trim().to_string()}pub fn read<T:FromStr>()->T{read_from(stdin())}pub fn read_token()->String{read_token_from(stdin())}pub fn read_ascii()->char{read_ascii_from(stdin())}pub fn read_from<R:Read,T:FromStr>(read:R)->T{read_token_from(read).parse().unwrap_or_else(|_|panic!("failed to parse a value"))}pub fn read_token_from<R:Read>(read:R)->String{read.bytes().flat_map(|x|x.map(|x|x as char)).skip_while(|x|x.is_whitespace()).take_while(|x|!x.is_whitespace()).collect()}pub fn read_ascii_from<R:Read>(read:R)->char{read.bytes().flat_map(|x|x.map(|x|x as char)).skip_while(|x|x.is_whitespace()).next().expect("failed to get a next character")}}}
use pcl::stdin::*;
use std::collections::VecDeque;
fn main() {
    let s = read_line();
    let q: usize = read();
    let mut reversed = false;
    let mut res: VecDeque<_> = s.chars().collect();
    for _ in 0..q {
        let t: i32 = read();
        if t == 1 {
            reversed = !reversed;
        } else {
            let f: i32 = read();
            let c: char = read_ascii();
            match (reversed, f) {
                (false, 1) => res.push_front(c),
                (false, 2) => res.push_back(c),
                (true, 1) => res.push_back(c),
                (true, 2) => res.push_front(c),
                _ => unreachable!(),
            }
        }
    }
    let iter = res.into_iter();
    let s: String = if !reversed {
        iter.collect()
    } else {
        iter.rev().collect()
    };
    println!("{}", s);
}