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)

function Main(input) {
	input = input.split("\n")
	var str = input[0];
	input.shift();
	var n = 1;
	for(var i=0;i<input.length;i++){
		var mem = input[i].split(" ");
		if(mem[0]==1){
			n *= -1;
		}else{
			switch(mem[1]-0){
				case 1:
				switch(n){
					case -1:
					str += mem[2];
					break;
					case 1:
					str = mem[2] += str;
					break;
				}
				break;
				case 2:
				switch(n){
					case 1:
					str += mem[2];
					break;
					case -1:
					str = mem[2] += str;
					break;
				}
				break;
			}
		}
     // console.log(str,n)
	}
if(n==1){
	console.log(str)
}else{
	console.log(str.split("").reverse().join(""))
}
}Main(require("fs").readFileSync("/dev/stdin", "utf8").trim());