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"
	"os"
	"strconv"
)

var sc, wr = bufio.NewScanner(os.Stdin), bufio.NewWriter(os.Stdout)

func scanString() string { sc.Scan(); return sc.Text() }
func scanRunes() []rune { return []rune(scanString()) }
func scanInt() int { a,_ := strconv.Atoi(scanString()); return a }
func scanInt64() int64 { a,_ := strconv.ParseInt(scanString(),10,64); return a }
func scanFloat64() float64 { a,_ := strconv.ParseFloat(scanString(),64); return a }

func scanInts(n int) []int {
	res := make([]int, n); for i := 0; i < n; i++ { res[i] = scanInt() }; return res
}

func debug(a ...interface{}) { fmt.Fprintln(os.Stderr, a...) }

func abs(a int) int { if a<0 { return -a }; return a }
func min(a,b int) int { if a<b { return a }; return b }
func max(a,b int) int { if a>b { return a }; return b }

//•*¨*•.¸¸♪Main•*¨*•.¸¸♪(　-ω-)ノ　(　・ω・)

func main() {
	defer wr.Flush()
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, 10000), 1001001)

	s := scanRunes()
	n := len(s)
	q := scanInt()
	ans := make([]rune,n+q*2)
	for i := 0; i < n; i++ {
		ans[i+q] = s[i]
	}
	l := q
	r := q+n-1
	now := true

	for i := 0; i < q; i++ {
		num := scanInt()
		if num == 1 {
			if now {
				now = false
			} else {
				now = true
			}
		} else {
			f := scanInt()
			c := scanRunes()
			if f == 1 {
				if now {
					l--
					ans[l] = c[0]
				} else {
					r++
					ans[r] = c[0]
				}
			} else {
				if now {
					r++
					ans[r] = c[0]
				} else {
					l--
					ans[l] = c[0]
				}
			}
		}
		// debug(string(ans),l,r,now)
	}

	ans = ans[l:r+1]
	for i := 0; i < len(ans); i++ {
		if now {
			// a[i] = ans[i]
			fmt.Fprint(wr, string(ans[i]))
		} else {
			// a[i] = ans[r-(i-l)]
			fmt.Fprint(wr, string(ans[len(ans)-1-i]))
		}
	}

	fmt.Fprintln(wr)
}