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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

//INCLUDE
//------------------------------------------
#include <iostream>
#include <functional>
#include <algorithm>
#include <math.h>
#include <iomanip>
#include <string>
#include <cstring>
#include <vector>
//DEFINE
//------------------------------------------
#define ll long long
#define ALLv(a)  (a).begin(),(a).end()
#define ALL(a,n)  (a),(a)+n
#define vi vector<long long>
#define vd vector<long double>
#define vs vector<string>
//CONST
//------------------------------------------
#define INF 1010000000000000017LL
#define MOD 1000000007LL
#define EPS 1e-12
#define PI 3.14159265358979323846
//REPEAT
//------------------------------------------
#define   FOR(i,m,n) for(ll (i)=(m);   (i)<(n);  (i)++)
#define   REP(i,n)   for(ll (i)=0;     (i)<(n);  (i)++)
#define  REPS(i,x)   for(ll (i)=1;     (i)<=(x); (i)++)
#define  RREP(i,x)   for(ll (i)=(x)-1; (i)>=0;   (i)--)
#define RREPS(i,x)   for(ll (i)=(x);   (i)> 0;   (i)--)
#define  WREP(i,in,j,jn) REP(i,in)REP(j,jn)
using namespace std;
//-------------------------------------------

int main(void){
  string S;cin>>S;
  ll Q;cin>>Q;
  ll q1,q2;char q3;
  REP(i,Q){
    cin>>q1;
    if(q1==1){
      reverse(ALLv(S));
    }else{
      cin>>q2>>q3;
      if(q2==1){
	S=q3+S;
      }else{
	S+=q3;
      }
    }
  }
  cout<<S<<endl;
  return 0;
}