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"
)

// +=での文字列連結はパフォーマンス的にあまりよろしくないらしい
// https://qiita.com/ruiu/items/2bb83b29baeae2433a79
// https://qiita.com/ono_matope/items/d5e70d8a9ff2b54d5c37
// https://drken1215.hatenablog.com/entry/2020/03/08/144800
// いくつか別解があるらしい
func main() {
	var S, t, l string
	var Q, rev int
	fmt.Scan(&S, &Q)
	for i := 0; i < Q; i++ {
		var T int
		fmt.Scan(&T)
		switch T {
		case 1:
			// 0と1を反転させる
			rev = 1 - rev
		case 2:
			var F int
			var C string
			fmt.Scan(&F, &C)
			F--
			if rev == 1 {
				F = 1 - F
			}
			if F == 0 {
				t = C + t
			} else {
				l = t + C
			}
		}
	}
	if rev == 1 {
		fmt.Println(Reverse(l) + Reverse(S) + Reverse(t))
	} else {
		fmt.Println(t + S + l)
	}
}

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)
}