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)

#![allow(non_snake_case)]
use proconio::input;
// use std::collections::HashMap;

fn main() {
    input! {
        S: String,
        Q: i64,
    }

    // 反転をまじめに扱おうとするとO(N)かかる。
    // 反転しているかどうかフラグを持って、それを最後に解決するようにすれば途中の計算はカットできる

    let mut is_normal = true;
    let mut head = Vec::new();
    let mut tail = Vec::new();

    for _i in 0..Q {
        input! {
            T: i64,
        }

        if T == 1 {
            // 逆転させるクエリ
            is_normal = !is_normal;
        } else {
            input! {
                F: i64,
                C: String,
            }

            if F == 1 {
                // 先頭に追加
                if is_normal {
                    head.push(C);
                } else {
                    tail.push(C);
                }
            } else {
                // 末尾に追加
                if is_normal {
                    tail.push(C);
                } else {
                    head.push(C);

                }
            }
        }
    }

    if is_normal {
        for s in head.iter().rev() {
            print!("{}", s.as_str());
        }

        print!("{}", S);

        for s in tail.iter() {
            print!("{}", s.as_str());
        }
    } else {
        for s in tail.iter().rev() {
            print!("{}", s);
        }

        let r = S.chars().rev().collect::<String>();
        print!("{}", r.as_str());


        for s in head.iter() {
            print!("{}", s.as_str());
        }
    }
    println!("");



}