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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"fmt"
)

func Reverse(s string) string {
	runes := []rune(s)
	for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
		runes[i], runes[j] = runes[j], runes[i]
	}
	return string(runes)
}

func main() {
	var text string

	fmt.Scan(&text)
	// fmt.Println(text)

	var Q int
	var T int
	var F int
	var C string
	fmt.Scan(&Q)

	var buf [500100]byte
	startcursor := 200050
	endcursor := startcursor + len(text)
	// // var buf [30]byte
	// buf[3] = 'a'
	copy(buf[startcursor:], []byte(text))
	// fmt.Printf("%v\n", text)
	// fmt.Printf("%v %v\n", startcursor, endcursor)
	// fmt.Printf("%v\n", buf)
	// fmt.Printf("%v\n", text[3])

	R := 0 // 0: original, 1: reversed
	for i := 0; i < Q; i++ {
		fmt.Scan(&T)
		// fmt.Printf("type %d\n", T)
		if T == 1 {
			// fmt.Printf("type1 %s\n", Reverse(text))
			// text = Reverse(text)
			R = (R + 1) % 2
		} else {
			fmt.Scan(&F, &C)
			if (F == 1 && R == 0) || (F == 2 && R == 1) {
				// text = C + text
				startcursor--
				buf[startcursor] = C[0]
			} else {
				buf[endcursor] = C[0]
				endcursor++
				// text = text + C
			}
			// fmt.Printf("%v, %v, %v, %v\n", Q, T, F, text)
		}
	}

	// fmt.Printf("%v\n", buf)
	// fmt.Printf("%v %v\n", startcursor, endcursor)
	text = string(buf[startcursor:endcursor])
	// fmt.Printf("%s\n", text)
	// fmt.Printf("%v\n", buf[startcursor:endcursor])
	if R == 1 {
		text = Reverse(text)
	}
	fmt.Println(text)
}