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;
use std::io::Read;
use std::iter::FromIterator;
use std::string::String;
fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut it = buf.split_whitespace();
    let s = it.next().unwrap();
    let q = it.next().unwrap().parse::<i64>().unwrap();
    let mut reversed = false;

    let mut deque = VecDeque::<char>::from_iter(s.chars());

    for _ in 0..q {
        let query = it.next().unwrap().parse::<i64>().unwrap();
        if query == 1 {
            reversed = !reversed;
        } else if query == 2 {
            let f = it.next().unwrap().parse::<i64>().unwrap();
            let c = it.next().unwrap().chars().next().unwrap();
            let push_front = if f == 1 { !reversed } else { reversed };
            if push_front {
                deque.push_front(c);
            } else {
                deque.push_back(c);
            }
        }
    }
    if reversed {

        let out = deque.into_iter().rev().collect::<String>();
        println!("{}", out);
    } else {
        let out = deque.into_iter().collect::<String>();
        println!("{}", out);
    }
}