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 C (Clang 3.8.0) (judge verdict: Accepted)

#include <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>

#define もし if(
#define そうでなくてもし }else if(
#define ならば ){
#define そうでないならば }else{
#define を実行する }

#define が ==
#define を =
#define とする

#define 文字列を読み込む(s) scanf("%s", s)
#define 整数を読み込む(x) int x; scanf("%d", &x)
#define 文字を読み込む(c) char c; scanf(" %c", &c)

#define 文字列を書き出す(s) printf("%s", (s))
#define 文字を書き出す(c) printf("%c", (c))
#define 改行する putchar(10)

#define 文字列の長さ(s) (strlen(s))

typedef int 整数;
typedef char 文字;

int main(void){
	文字 s[500001];
	文字列を読み込む(s+200000);
	整数 l を 200000 とする;
	整数 r を 200000+文字列の長さ(s+200000)とする;
	整数 f を 1 とする;

	整数を読み込む(q);
	整数 i;
	for(i=0; i<q; i++){
		整数を読み込む(Q);
		もし Q が 1 ならば
			もし f が 1 ならば
				f を 2 とする;
			そうでないならば
				f を 1 とする;
			を実行する
		そうでないならば
			整数を読み込む(F);
			文字を読み込む(c);
			もし F が f ならば
				s[--l] を c とする;
			そうでないならば
				s[r++] を c とする;
			を実行する
		を実行する
	}
	もし f が 1 ならば
		s[r] を 0 とする;
		文字列を書き出す(s+l);
		改行する;
	そうでないならば
		for(i=r-1; i>=l; i--){
			文字を書き出す(s[i]);
		}
		改行する;
	を実行する
}