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: Wrong Answer)

#include<stdio.h>

typedef struct{
    int kind;
    int f;
    char c;
    int rev;
}query;


int main(void){
    char i_s[100001];
    scanf("%s",i_s);
    int q_n;
    scanf("%d",&q_n);
    query q[q_n];
    for(int i = 0;i < q_n;i++){
        scanf("%d",&q[i].kind);
        if(q[i].kind == 2)scanf("%d %c",&q[i].f,&q[i].c);
        q[i].rev = 0;
    }
    for(int i = 0;i < q_n;i++){
        if(q[i].kind == 1){
            for(int j = i-1;j >= 0;j--){
                if(q[j].kind = 2){
                    q[j].rev = (q[j].rev?0:1);
                }
            }
        }
    }
    for(int i = q_n-1;i >= 0;i--)if(q[i].kind == 2){
        if(q[i].f == 1 && !(q[i].rev) || q[i].f == 2 && (q[i].rev))printf("%c",q[i].c);
    }
    printf("%s",i_s);
    for(int i = 0;i < q_n;i++)if(q[i].kind == 2){
        if(q[i].f == 2 && !(q[i].rev) || q[i].f == 1 && (q[i].rev))printf("%c",q[i].c);
    }
    return 0;
}