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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
 
#define N 100000

int q;
char s[3*N];

int main(){
    scanf("%s",s);
    scanf("%d",&q);
    int slen = strlen(s);
    int t, sMin=2*N, sMax=sMin+slen-1, reverse=0; 
    char f, c, sMod[5*N];

    strncpy(sMod+sMin, s, slen);
    for (int i = 0; i < q; i++){
        scanf("%d",&t);
        if(t==1){
            reverse = (reverse+1)&1;
        }
        else{
            scanf("%d %c", &f, &c);    //　型がごちゃごちゃ
            if(f==1){
            //先頭にc
                if(reverse==0){
                    sMin--;
                    sMod[sMin] = c;
                }
                else{
                    sMax++;
                    sMod[sMax] = c;
                }
            }
            else{
            //末尾にc
                if(reverse==0){
                    sMax++;
                    sMod[sMax] = c;
                }
                else{
                    sMin--;
                    sMod[sMin] = c;
                }
            }
        }
    }
    if(reverse==0){
        strncpy(s, sMod+sMin, sMax-sMin+1 );
    }
    else{
        int cnt = 0;
        for(int i = sMax; i>=sMin; i--){
            s[cnt] = sMod[i];
            cnt++;
        }
    }
    slen = strlen(s);
    s[slen] = '\0';
        
    printf("%s\n", s);
    return 0;
}