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>
#include <ext/pb_ds/assoc_container.hpp> 
using namespace __gnu_pbds;
using namespace std;
#define F first
#define S second
#define pb push_back
#define powerof2 ()
using ll = long long;
const ll M = 1000000007;
const ll infi = 1000000000000000;
const ll md = 998244353;
using vl = vector<ll>;
using pll =  pair<ll,ll>;
typedef tree<int,null_type,less<int>,rb_tree_tag, tree_order_statistics_node_update> is;
ll power(ll x,ll y){
    if(y==0)
    return 1;
    ll res=power(x,y/2);
    res=(res*res)%M;
    if(y&(ll)1)
    res=(res*x)%M;
    return res;
}


/*int n,ma=0;
vector<vector<int>> gr;
vector<int> color;
void dfs(int c,int p)
{
    vector<int> v;
    int j;
    for(j=1;j<gr[c].size()+2;j++)
    {
        if(j==color[c]||j==color[p])
        continue;
        //cout<<j<<" ";
        v.pb(j);
    }
    //cout<<endl;
    j=0;
    for(int i=0;i<gr[c].size();i++)
    {
        if(gr[c][i]!=p)
        {
            color[gr[c][i]]=v[j];j++;
            dfs(gr[c][i],c);
        }
    }
}*/
void solve(){
    string s;
    int q;
    cin>>s>>q;
    string f="",b="";
    int c=0;
    while(q--)
    {
        int ty;cin>>ty;
        if(ty==1)
        {
            c++;
        }
        else
        {
            int p;char ch;
            cin>>p>>ch;
            if(p==1)
            {
                if(c&1)
                b.pb(ch);
                else
                f.pb(ch);
            }
            else
            {
                if(c&1)
                f.pb(ch);
                else
                b.pb(ch);
            }
        }
    }
    reverse(f.begin(),f.end());
    s=f+s+b;
    if(c&1)
    reverse(s.begin(),s.end());
    cout<<s;
}
 
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    // int t;
    // cin>>t;
    // while(t--)
    solve();
    return 0;
}