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 (
	"bufio"
	"fmt"
	"io"
	"os"
	"strconv"
)

func main() {
	resolve(parseProblem(os.Stdin))
}

func parseProblem(r io.Reader) (string, chan query) {
	sc := bufio.NewScanner(r)
	sc.Split(bufio.ScanWords) // bufio.ScanLines
	const (
		initialBufSize = 100000
		maxBufSize     = 100000
	)
	buf := make([]byte, initialBufSize)
	sc.Buffer(buf, maxBufSize)

	sc.Scan()
	s := sc.Text()
	n := scanint(sc)

	ch := make(chan query)
	go func() {
		for i := 0; i < n; i++ {
			ch <- parseQuery(sc)
		}
		close(ch)
	}()
	return s, ch
}

const (
	Switch = 1 + iota
	Append
)

type query struct {
	cmd int
	sub int
	c   string
}

func parseQuery(sc *bufio.Scanner) query {
	cmd := scanint(sc)
	if cmd == Switch {
		return query{cmd, 0, ""}
	}
	sub := scanint(sc)
	sc.Scan()
	t := sc.Text()
	return query{Append, sub, t}
}

func scanint(sc *bufio.Scanner) int {
	sc.Scan()
	i, _ := strconv.Atoi(sc.Text())
	return int(i)
}

func resolve(base string, queries chan query) {
	state := 0
	fixies := [2][]string{}
	for q := range queries {
		if q.cmd == Switch {
			state += 1
		} else {
			// state 0; 1 -> 1(prefix), 2 -> 0 (suffix)
			fixies[(state+q.sub)%2] = append(fixies[(state+q.sub)%2], q.c)
		}
	}

	normal := state%2 == 0
	prefix := fixies[(state+1)%2]
	suffix := fixies[(state+2)%2]

	for i := len(prefix); i > 0; i-- {
		fmt.Printf("%s", prefix[i-1])
	}
	if normal {
		fmt.Printf("%s", base)
	} else {
		runes := []rune(base)
		for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
			runes[i], runes[j] = runes[j], runes[i]
		}
		fmt.Printf("%s", string(runes))
	}

	for i := 0; i < len(suffix); i++ {
		fmt.Printf("%s", suffix[i])
	}
}