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: Wrong Answer)

fn main() {
    let mut s = read::<String>();
    let qnum = read::<u32>();

    let mut rev = false;
    let mut qs = vec![];
    for _ in 0..qnum {
        let line = read_vec::<String>();
        if line[0] == "1" {
            rev = !rev;
        }
        qs.push(line);
    }

    if rev {
        s = s.chars().rev().collect::<String>();
    }

    rev = false;
    for q in qs {
        let t: u32 = q[0].parse().ok().unwrap();
        if t == 1 {
            rev = !rev;
        } else {
            let f: u32 = q[1].parse().ok().unwrap();
            let c: char = q[2].clone().chars().collect::<Vec<_>>()[0];
            if f == 1 {
                if !rev {
                    s.insert(0, c);
                } else {
                    s.push(c);
                }
            } else {
                if !rev {
                    s.push(c);
                } else {
                    s.insert(0, c);
                }
            }
        }
    }
    if rev {
        s = s.chars().rev().collect::<String>();
    }
    println!("{}", s);
}

#[allow(dead_code)]
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}
#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
#[allow(dead_code)]
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
#[allow(dead_code)]
fn yn(result: bool) {
    if result {
        println!("Yes");
    } else {
        println!("No");
    }
}