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: Wrong Answer)

use proconio::input;

fn main() {
    input! {
        mut s: String,
        q: usize,
    }
    let mut queries = vec![Vec::new(); q];
    for i in 0..q {
        input! {
            head: char
        }
        queries[i].push(head);
        if head == '2' {
            input! {
                x: char,
                y: char
            }
            queries[i].push(x);
            queries[i].push(y);
        }
    }

    let mut now = 0;
    let mut former = String::new();
    while queries[now][0] == '2' {
        if queries[now][1] == '1' {
            former.push(queries[now][2]);
        } else {
            s.push(queries[now][2]);
        }
        now += 1;
        if now == q {
            break;
        }
    }
    let mut s = format!("{}{}", former.chars().rev().collect::<String>(), s);

    if now == q {
        println!("{}",s)
    } else {
        let mut rev = vec![0; q-now];
        rev[0] = 1;
        for i in (1..q-now).rev() {
            if queries[i][0] == '1' {
                rev.push(rev[i-1]+1);
            } else {
                rev.push(rev[i-1]);
            }
        }
        rev.reverse();

        if rev[0]%2 == 1 {
            s = s.chars().rev().collect::<String>();
        }

        let mut former = String::new();
        for i in 1..q-now {
            if queries[i+now][0] == '2' {
                if queries[i+now][1] == '2' && rev[i]%2 == 0 || queries[i+now][1] == '1' && rev[i]%2 == 1 {
                    former.push(queries[i+now][2]);
                } else {
                    s.push(queries[i+now][2]);
                }
            }
        }
        let ans = format!("{}{}", former.chars().rev().collect::<String>(), s);
        println!("{}",ans)
    }
}