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 (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

func main() {
	nextReader = newScanner()
	s := nextString()
	q := nextInt()
	cnt := 0
	for i := 0; i < q; i++ {
		t := nextInt()
		if t == 1 {
			cnt++
		}
		if t == 2 {
			f, c := nextInt(), nextString()
			if f == 1 {
				if cnt%2 == 0 {
					s = c + s
				} else {
					s = s + c
				}
			}
			if f == 2 {
				if cnt%2 == 0 {
					s = s + c
				} else {
					s = c + s
				}
			}
		}
	}
	if cnt%2 == 0 {
		fmt.Println(s)
	} else {
		fmt.Println(reverse(s))
	}
}

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

// ---------------------------------------------------------------
// I/O
// ---------------------------------------------------------------
var nextReader func() string

func newScanner() func() string {
	sc := bufio.NewScanner(os.Stdin)
	sc.Buffer(make([]byte, 1024), int(1e11))
	sc.Split(bufio.ScanWords)
	return func() string {
		sc.Scan()
		return sc.Text()
	}
}

func nextString() string { return nextReader() }

func nextInt() int { n, _ := strconv.Atoi(nextReader()); return n }

func nextInts(size int) []int {
	ns := make([]int, size)
	for i := 0; i < size; i++ {
		ns[i] = nextInt()
	}
	return ns
}