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: Wrong Answer)

package main

import (
	"fmt"
	"strings"
)

var ans string
var s string
var q int
var t, f int
var c string

func main() {
	fmt.Scan(&s)
	fmt.Scan(&q)
	fronts := make([]rune, 0, 200000)
	backs := make([]rune, 0, 200000)

	rev := false
	for i := 0; i < q; i++ {
		fmt.Scan(&t)
		switch t {
		case 1:
			rev = !rev
		case 2:
			fmt.Scan(&f, &c)
			head := (f == 1)
			if rev {
				head = !head
			}
			if head {
				fronts = append(fronts, []rune(c)[0])
			} else {
				backs = append(backs, []rune(c)[0])
			}
		default:
			panic("err")
		}
	}

	if rev {
		ans = strings.Join([]string{
			string(reverse(backs)), s, string(fronts),
		}, "")
	} else {
		ans = strings.Join([]string{
			string(reverse(fronts)), s, string(backs),
		}, "")

	}
	fmt.Println(ans)
}

func reverse(ss []rune) []rune {
	o := make([]rune, len(ss))
	i := len(ss)
	for _, s := range ss {
		i--
		o[i] = s
	}
	return o
}