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: Runtime Error)

use std::iter::FromIterator;
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()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|s| s.parse().ok().unwrap())
        .collect()
}

fn main() {
    let s: String = read();
    let q: usize = read();

    let mut buf = [0; 200000];
    let mut si = 100000;
    let mut ei = 100000;

    buf[si] = s.chars().collect::<Vec<_>>()[0] as u8;

    let mut is_reverse = false;
    for _ in 0..q {
        let input = read_vec::<String>();
        if input[0] == "1" {
            is_reverse = !is_reverse;
        } else if input[1] == "1" {
            if !is_reverse {
                si -= 1;
                buf[si] = input[2].chars().collect::<Vec<_>>()[0] as u8;
            } else {
                ei += 1;
                buf[ei] = input[2].chars().collect::<Vec<_>>()[0] as u8;
            }
        } else {
            if !is_reverse {
                ei += 1;
                buf[ei] = input[2].chars().collect::<Vec<_>>()[0] as u8;
            } else {
                si -= 1;
                buf[si] = input[2].chars().collect::<Vec<_>>()[0] as u8;
            }
        }
    }

    let mut ans: Vec<char> = Vec::new();
    if !is_reverse {
        for i in si..(ei + 1) {
            ans.push(buf[i] as char);
        }
    } else {
        for i in (si..(ei + 1)).rev() {
            ans.push(buf[i] as char);
        }
    }

    println!("{}", String::from_iter(ans));
}