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: Time Limit Exceeded)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

func addTail(s *string, t string) {
	*s += t
}

func addHead(s *string, t string) {
	*s = t + (*s)
}

func reverseString(s string) string {
	n := len(s)
	bytes := []byte(s)
	for i := 0; i < n/2; i++ {
		bytes[i], bytes[n-1-i] = bytes[n-1-i], bytes[i]
	}
	return string(bytes)
}

func main() {
	sc.Split(bufio.ScanWords)
	// s := nextMega()
	// var s string
	// fmt.Scan(&s)
	sc.Buffer(make([]byte, 100001), 100001*100)
	s := next()
	q := nextInt()
	rev := false

	for i := 0; i < q; i++ {
		t := nextInt()
		switch t {
		case 1:
			rev = !rev
		case 2:
			f, c := nextInt(), next()
			if (!rev && f == 1) || (rev && f == 2) {
				addHead(&s, c)
			} else {
				addTail(&s, c)
			}
		}
	}

	if !rev {
		puts(s)
	} else {
		puts(reverseString(s))
	}
	wt.Flush()
}

var (
	sc  = bufio.NewScanner(os.Stdin)
	rdr = bufio.NewReaderSize(os.Stdin, 1000000)
	wt  = bufio.NewWriter(os.Stdout)
)

func next() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	i, _ := strconv.Atoi(next())
	return i
}

func nextMega() string {
	buf := make([]byte, 0, 1000000)
	for {
		l, p, _ := rdr.ReadLine()
		buf = append(buf, l...)
		if !p {
			break
		}
	}
	return string(buf)
}

func puts(a ...interface{}) {
	fmt.Fprintln(wt, a...)
}