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 <math.h>

#define rep(i, n) for(i=0;i<(int)(n);i++)
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
#define bit(n) (1LL<<(n))
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return (a*b)/gcd(a,b);}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

void rev_string(char str[])
{
    int i;
    int len = strlen(str);
    for (i = 0; i < len / 2; i++) {
        char temp = str[i];
        str[i] = str[len-i-1];
        str[len-i-1] = temp;
    }
}

int main(void){
  char s[500001];
  int f[200001];
  char c[200001]={'\0'};
  int rev[200001];
  int q,i,j,revMode=0,start=250000,len=0;
  
  scanf("%s", &s[start]);
  scanf("%d", &q);
  len = strlen(&s[start]);
  for(i=0;i<q;i++){
    scanf("%d", &j);
    switch(j){
      case 1:
        revMode += 1;
        revMode %= 2;
        break;
      case 2:
        scanf("%d %c", &f[i], &c[i]);
        if(revMode == 1){
          f[i] %= 2;
          f[i] += 1;//hanten
        }
        if(f[i] == 1){
          //atama
          f[i] = start-1;
          start--;
        }else{
          //ketsu
          f[i] = start+len;
        }
        len++;
        
        break;
    }
  }
  for(i=0;i<q;i++){
    //create string
    s[f[i]] = c[i];
  }
  if(revMode == 0) printf("%s", &s[start]);
  else{
    //printf("%s", &s[start]);
    for(i=start+strlen(&s[start])-1;i>=start;i--){
      printf("%c", s[i]);
    }
  }
  
  return 0;
}