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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdint.h>
#include<stdbool.h>
#include<tgmath.h>
#include<time.h>
#define ll long long
#define min(a,b) ((a)>(b)?(b):(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define miin(a,b,c) min((a), min((b), (c)))
#define maax(a,b,c) max((a), max((b), (c)))

char s[1111111] = {}, c;
int q, a, b, i = 500000, j, nyan = 0;

int main(){
    j = i;while((c=getchar()) != 10){s[j] = c;j++;}j--;
    scanf("%d", &q);
    while(q--){
        scanf("%d", &a);
        if(a == 1){
            nyan ^= 1;
        }else {
            scanf("%d %c",&b, &c);
            if(b == 1){
                if(nyan == 0){
                    i--;
                    s[i] = c;
                }else {
                    j++;
                    s[j] = c;
                }
            }else {
                if(nyan == 0){
                    j++;
                    s[j] = c;
                }else {
                    i--;
                    s[i] = c;
                }
            }
        }
    }
    if(nyan == 0){
        for(int k=i;k<=j;k++)putchar(s[k]);
    }else{
        for(int k=j;k>=i;k--)putchar(s[k]);
    }
    return 0;
}