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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

"use strict";
    
const main = arg => {
    arg = arg.trim().split("\n");
    let S = arg[0].split("");
    const N = parseInt(arg[1]);
    const Q = arg.slice(2, N + 2);
    
    let prev = [];
    let next = [];
    
    let flip = false;
    
    for(let i=0; i<N; i++) {
        const split = Q[i].split(" ");
        // Ti = 1の時、反転
        if(split.length === 1) {
            if(flip) {
                flip = false
                let temp = prev;
                prev = next;
                next = temp;
            } else {
                flip = true;
                let temp = prev;
                prev = next;
                next = temp;
            }
        } else {
            if(split[1] === "1") {
                if(!flip) {
                    prev.push(split[2]);
                } else {
                    next.push(split[2]);
                }
            } else {
                if(flip) {
                    next.push(split[2]);
                } else {
                    prev.push(split[2]);
                }
            }
        }
        
        // console.log("flip;" + flip)
        // console.log("prev:" + prev);
        // console.log("next:" + next);
        // console.log("---")
    }
    

    
    console.log(prev.join("") + S + next.join(""));
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));