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 (
	"fmt"
	"strings"
)

func swap(s string) string {
	if len(s) == 1 {
		return s
	}
	return s[len(s)-1:] + s[1:len(s)-1] + s[:1]
}

func Reverse(s string) string {
	runes := []rune(s)

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

	return string(runes)
}

func main() {
	var s string
	var n int
	fmt.Scan(&s, &n)

	var q, f int
	var c string
	var reverse bool
	front := make([]string, 0)
	back := make([]string, 0)
	for i := 0; i < n; i++ {
		fmt.Scan(&q)
		if q == 1 {
			if reverse {
				reverse = false
			} else {
				reverse = true
			}
		} else {
			fmt.Scan(&f, &c)
			if f == 1 {
				if reverse {
					back = append(back, c)
				} else {
					front = append([]string{c}, front...)
				}
			} else {
				if reverse {
					front = append([]string{c}, front...)
				} else {
					back = append(back, c)
				}
			}
		}
	}

	s = strings.Join(front, "") + s + strings.Join(back, "")
	if reverse {
		s = Reverse(s)
	}
	fmt.Println(s)
}