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<bits/stdc++.h>

#define bb      begin()
#define ee      end()
#define ff      first
#define ss      second
#define pb      push_back
#define pf      push_front
#define popb    pop_back()
#define popf    pop_front()
#define mem(a , v)    memset(a , v , sizeof(a))

typedef long long ll;
typedef long double ld;

using namespace std;

const int BIG = 1e5+9;
const int INF = 1e9+9;
const ll BINF = 1e18+9;
const double SML = (1e-7);

int q,t,f,i;
char c;
string s;
list<char> l;
list<char>::const_iterator it;
bool rev=false;

int main()
{
    //freopen("output.txt" , "w" , stdout);
    //freopen("input.txt" , "r" , stdin);
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin>>s;
    for(i=0 ; i<s.size() ; i++)
    {
        l.pb(s[i]);
    }
    cin>>q;
    for(i=0 ; i<q ; i++)
    {
        cin>>t;
        if(t==1)
        {
            if(rev)
            {
                rev=false;
            }
            else
            {
                rev=true;
            }
        }
        else
        {
            cin>>f>>c;
            if(f==1)
            {
                if(!rev)
                {
                    l.pf(c);
                }
                else
                {
                    l.pb(c);
                }
            }
            else
            {
                if(!rev)
                {
                    l.pb(c);
                }
                else
                {
                    l.pf(c);
                }
            }
        }
    }
    if(rev)
    {
        reverse(l.bb , l.ee);
    }
    for(it=l.bb ; it!=l.ee ; it++)
    {
        cout<<*it;
    }
    return 0;
}