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 string = read();
    let yosoku_string_len = string.len() + 2 * 100000;
    let mut appended_to_start = Vec::with_capacity(yosoku_string_len / 2);
    let mut appended_to_end = Vec::with_capacity(yosoku_string_len / 2);
    let query_cnt = read().parse::<u64>().unwrap();
    let mut reversed = false;
    for _ in 0..query_cnt {
        let query = read_as_vec();
        let op = query[0].parse::<u64>().unwrap();
        if op == 1 {
            reversed = !reversed;
        } else { // op == 2
            let append_to_start = query[1].parse::<u64>().unwrap() == 1;
            let ch = query[2].chars().nth(0).unwrap();
            if reversed {
                // reversed ops
                if append_to_start {
                    appended_to_end.push(ch);
                } else {
                    appended_to_start.push(ch);
                }
            } else {
                // nature order
                if append_to_start {
                    appended_to_start.push(ch);
                } else {
                    appended_to_end.push(ch);
                }
            }
        }
    }

    let start = if !reversed { &appended_to_start } else { &appended_to_end };
    let end = if !reversed { &appended_to_end } else { &appended_to_start };

    let result_len = start.len() + string.len() + end.len();
    let mut build = String::with_capacity(result_len);
    for i in (0..start.len()).rev() {
        build.push(start[i]);
    }
    build.push_str(&string);
    for i in 0..end.len() {
        build.push(end[i]);
    }

    println!("{}", build);
}
fn read() -> String {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();
    line.trim().to_string()
}
fn read_as_vec() -> std::vec::Vec<String> {
    read().split(' ').map(|s| s.to_string()).collect()
}