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: Accepted)

use std::collections::VecDeque;
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(|e| e.parse().ok().unwrap()).collect()
}

fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
	(0..n).map(|_| read_vec()).collect()
}

pub trait LexicalPermutation {
	/// Return `true` if the slice was permuted, `false` if it is already
	/// at the last ordered permutation.
	fn next_permutation(&mut self) -> bool;
}
impl<T> LexicalPermutation for [T] where T: Ord {
	/// Original author in Rust: Thomas Backman <serenity@exscape.org>
	fn next_permutation(&mut self) -> bool {
		// These cases only have 1 permutation each, so we can't do anything.
		if self.len() < 2 { return false; }

		// Step 1: Identify the longest, rightmost weakly decreasing part of the vector
		let mut i = self.len() - 1;
		while i > 0 && self[i-1] >= self[i] {
			i -= 1;
		}

		// If that is the entire vector, this is the last-ordered permutation.
		if i == 0 {
			return false;
		}

		// Step 2: Find the rightmost element larger than the pivot (i-1)
		let mut j = self.len() - 1;
		while j >= i && self[j] <= self[i-1]  {
			j -= 1;
		}

		// Step 3: Swap that element with the pivot
		self.swap(j, i-1);

		// Step 4: Reverse the (previously) weakly decreasing part
		self[i..].reverse();

		true
	}
}

fn main() {
	let s: String = read();
	let q: usize = read();
	let queries: Vec<Vec<String>> = read_vec2(q as u32);

	let mut s: VecDeque<char> = s.chars().collect();
	let mut reverse_flag: bool = false;

	for i in 0..queries.len() {
		if queries[i][0] == String::from("1") {
			if s.len() > 1 {
				reverse_flag = !reverse_flag
			}
		}  else {
			let target = queries[i][2].chars().nth(0).unwrap();
			if queries[i][1] == String::from("1") {
				if reverse_flag { s.push_back(target) }
				else { s.push_front(target) }
			} else {
				if reverse_flag { s.push_front(target) }
				else { s.push_back(target)}
			}
		}
	}

	if reverse_flag {
		for i in 0..s.len() {
			print!("{}", s[s.len()-1 - i])
		}
		println!();
	} else {
		for si in s {
			print!("{}", si);
		}
		println!();
	}
}