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"
)

var sc *bufio.Scanner

func init() {
	// 100MB
	bufsize := 100 * 1024 * 1024
	sc = bufio.NewScanner(os.Stdin)
	sc.Buffer(make([]byte, bufio.MaxScanTokenSize), bufsize)
	sc.Split(bufio.ScanWords)
}
func readInt() int              { return readIntArray(1)[0] }
func readInt2() (int, int)      { a := readIntArray(2); return a[0], a[1] }
func readInt3() (int, int, int) { a := readIntArray(3); return a[0], a[1], a[2] }
func readIntArray(n int) []int {
	s := readStringArray(n)
	a := make([]int, n)
	for i := range a {
		a[i], _ = strconv.Atoi(s[i])
	}
	return a
}
func readString() string            { return readStringArray(1)[0] }
func readString2() (string, string) { a := readStringArray(2); return a[0], a[1] }
func readStringArray(n int) []string {
	a := make([]string, n)
	for i := 0; i < n; i++ {
		sc.Scan()
		a[i] = sc.Text()
	}
	return a
}
func max(ns ...int) int {
	max := ns[0]
	for _, n := range ns {
		if n > max {
			max = n
		}
	}
	return max
}
func min(ns ...int) int {
	min := ns[0]
	for _, n := range ns {
		if n < min {
			min = n
		}
	}
	return min
}
func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

type Query struct {
	T int
	F int
	C rune
}

func (q *Query) Process(s []rune, rev bool) ([]rune, bool) {
	if q.T == 1 {
		//	fmt.Printf("(REV) s: %s, rev: %t -> (%s, %t)\n", s, rev, s, !rev)
		return s, !rev
	}

	//befores := s
	if q.F == 1 {
		if rev {
			s = append(s, q.C)
		} else {
			s, s[0] = append(s[:1], s[0:]...), q.C
		}
	} else {
		if rev {
			s, s[0] = append(s[:1], s[0:]...), q.C
		} else {
			s = append(s, q.C)
		}
	}

	//fmt.Printf("(APD) s: %s, rev: %t -> (%s, %t)\n", befores, rev, s, rev)
	return s, rev
}

func main() {
	S := readString()
	Q := readInt()
	qs := make([]Query, Q)
	for i := 0; i < Q; i++ {
		t := readInt()
		qs[i].T = t
		if t == 1 {
			continue
		}

		f := readInt()
		c := []rune(readString())[0]
		qs[i].F = f
		qs[i].C = c
	}

	var rev bool
	r := make([]rune, 0, 1001001)
	r = append(r, []rune(S)...)
	for _, q := range qs {
		r, rev = q.Process(r, rev)
	}

	if rev {
		fmt.Println(string(Reverse(r)))
		return
	}
	fmt.Println(string(r))
}

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