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: Accepted)

#include <iostream>
#include <string>
#include <math.h>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <set>
#include <queue>
#include <deque>
#include <map>
using namespace std;

#define mod 1000000007
#define ten5 100005
#define ten52 200005
#define ten6 1000005
#define PI 3.1415926
#define pb(x) push_back(x)
#define all(x) x.begin(),x.end()

typedef long long int ll;

//ll num[50];
//ll num2[50];
ll num[10];
//set<ll> sll;
//map<string,ll> mp,mp1;
//map<string,ll>::iterator iter;
deque<char> deq;

int main(void)
{
    long long int sum=0,range=0,ceiling;
    double ans=0.0,d,d1,d2;
    long long int m,n,o,i,j,k=1,money;
    long long int a,a1=0,a2=0,a3=0,ar=1,as=1,at1,inv=0;
    string s;
    char c,c1;
    a=a1=a2=a3=0;
    //vector <int> v;
    //float f,g,h;
    //int c,d,big=0;
    cin>>s>>n;
    for(i=0;i<s.size();i++) deq.pb(s[i]);
    for(i=0;i<n;i++)
    {
        cin>>a;
        if(a==1) inv=inv^1;     //inverse
        else
        {
            cin>>a1>>c1;
            if(inv==0 && a1==1) deq.push_front(c1);
            else if(inv==1 && a1==2) deq.push_front(c1);
            else if(inv==0 && a1==2) deq.push_back(c1);
            else if(inv==1 && a1==1) deq.push_back(c1);
        }
    }
    k=deq.size();
    for(i=0;i<k;i++)
    {
        if(inv==0) cout<<deq[i];
        else cout<<deq[k-1-i];
    }
    return 0;
}