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 (1.42.0) (judge verdict: Accepted)

// -*- coding:utf-8-unix -*-

use proconio::input;
// use proconio::marker::Chars;

fn main() {
    input! {
        s: String,
        q: u64,
    }

    let mut vs: Vec<char> = s.chars().collect();
    let mut is_reversed = false;
    let mut vhead: Vec<char> = vec!();
    let mut vtail: Vec<char> = vec!();
    for _i in 0..q {
        input! {
            c0: char,
        }

        // println!("{:?}", query);
        if c0 == '1' {
            is_reversed = !is_reversed;
        } else {
            input! {
                c1: char,
                c2: char,
            }
            if is_reversed {
                if c1 == '1' {
                    vtail.push(c2);
                } else {
                    vhead.push(c2);
                }
            } else {
                if c1 == '1' {
                    vhead.push(c2);
                } else {
                    vtail.push(c2);
                }
            }
        }
    }

    vhead.reverse();
    vhead.append(&mut vs);
    let mut vs =vhead;
    vs.append(&mut vtail);
    if is_reversed {
        vs.reverse();
    }
    println!("{}", vs.into_iter().collect::<String>());
}