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 (
	"fmt"
)

func main() {
	rev := false
	s1 := make([]rune, 300000)
	s2 := make([]rune, 200000)

	var s string
	fmt.Scanf("%s", &s)
	r := []rune(s)
	for i := 0; i < len(r); i++ {
		s1[i] = r[i]
	}
	len1, len2 := len(s), 0

	var Q int
	fmt.Scanf("%d", &Q)
	for Q > 0 {
		Q--
		var t int
		fmt.Scanf("%d", &t)
		if t == 1 {
			rev = !rev
			continue
		}
		var f int
		var c string
		fmt.Scanf("%d %s", &f, &c)
		if (f == 1 && rev) || (f == 2 && !rev) {
			s1[len1] = []rune(c)[0]
			len1++
		} else {
			s2[len2] = []rune(c)[0]
			len2++
		}
	}
	s1 = s1[:len1]
	s2 = s2[:len2]
	if rev {
		s1, s2 = s2, s1
	}
	fmt.Printf("%s\n", string(append(reverse(s2), s1...)))
}

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