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::io::*;
 
fn read(s: &mut StdinLock) -> usize {
    s.by_ref().bytes().map(|c| c.unwrap() as char)
        .take_while(|c| !c.is_whitespace())
        .skip_while(|c| c.is_whitespace())
        .fold(0, |a, x| (x as u8 - b'0') as usize + a * 10)
}
fn read_line_char(s: &mut StdinLock) -> Vec<char> {
    s.by_ref().bytes().map(|c| c.unwrap() as char)
        .take_while(|c| !c.is_whitespace())
        .skip_while(|c| c.is_whitespace())
        .collect()
}
fn read_char(s: &mut StdinLock) -> char {
    s.by_ref().bytes().map(|c| c.unwrap() as char)
        .take_while(|c| !c.is_whitespace())
        .skip_while(|c| c.is_whitespace())
        .collect::<Vec<char>>()[0]
}
 
fn main() {
    let s = stdin();
    let mut s = s.lock();
    let mut answer = read_line_char(&mut s);
    let q = read(&mut s);
    let mut sw = false;
    for _ in 0..q {
        if read_char(&mut s) == '1' {
            sw = !sw;
        } else {
            let key = read_char(&mut s);
            if (key == '1' && !sw) || (key =='2' && sw) {
                answer.insert(0, read_char(&mut s));
            } else {
                answer.push(read_char(&mut s));
            }
        }
    }
    if sw {
        println!("{}", answer.iter().rev().map(|x| *x).collect::<String>());
    } else {
        println!("{}", answer.iter().map(|x| *x).collect::<String>());
    }
}