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<bits/stdc++.h>
#include<cmath>
#define pb push_back
#define ld long double
#define mp make_pair
#define vl vector<ll> 
#define vd vector<double>
#define vld vector<long double>
#define ll long long int
#define pl pair<ll, ll>
#define all(a) a.begin(), a.end()
#define forr(i, n) for(ll i=0; i<n; i++) 
#define forr1(i, n) for(ll i=1; i<=n; i++)
using namespace std;
const ld PI =3.1415926535897923846;
const ll MOD = 1e9+7;
void solve()
{
    string str;
    cin>>str;
    int q;
    scanf("%d", &q);
    bool chance=false;
    forr(i, q)
    {
        int num;
        scanf("%d", &num);
        if(num==1)
        {
            chance=!chance;
            continue;
        }
        else
        {
            ll a1; char ch;
            cin>>a1>>ch;
            if(a1==1)
            {
                if(chance==false)
                {
                    str=ch+str;
                }
                else
                {
                    str=str+ch;
                }
            }
            if(a1==2)
            {
                if(chance!=false)
                {
                    str=ch+str;
                }
                else
                {
                    str=str+ch;
                }
            }
        }
    }
    if(!chance)
    cout<<str;
    else
    {
        for(ll i=str.size()-1; i>=0; i--)
        cout<<str[i];
    }
}
int main()
{
    ll test=1;
    //cin>>test;
    while(test--)
        solve();
}