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: Wrong Answer)

#include<bits/stdc++.h>
#include<algorithm>
#include<string>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

#define FASTER ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define pi acos(-1.0)

#define sf(a) scanf("%lld",&a)
#define sff(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define pf(a) printf("%lld\n",a)

#define stll(a) stoll(a,nullptr,10)
#define PII pair<ll,ll>
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define EL printf("\n")
#define ff first
#define ss second

#define mem(ar,val) memset(ar,val,sizeof(ar))
#define ssort(ar) sort(ar.begin(),ar.end())
#define csort(ar,comp) sort(ar.begin(),ar.end(),comp)

#define bug1(a) cout<<"test "<<a<<endl;
#define bug2(a,b) cout<<"test "<<a<<" "<<b<<endl;
#define bug3(a,b,c) cout<<"test "<<a<<" "<<b<<" "<<c<<endl;
#define bug4(a,b,c,d) cout<<"test "<<a<<" "<<b<<" "<<c<<" "<<d<<endl;

typedef long long ll;
typedef unsigned long long ull;
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;

int main()
{
    string s,s1,a="",b="";

    ll q,t,f,cnt=0;

    cin>>s>>q;

    while(q--)
    {
        cin>>t;
        if(t==1)   cnt++;
        else
        {
            cin>>f>>s1;
            if(f==1 && cnt%2==0)
                a+=s1;
            else if(  f==1 &&cnt%2!=0)
                b+=s1;
            else if(f==2 && cnt%2==0)
                b+=s1;
            else if(f==2 && cnt%2!=0)
                a+=s1;
        }
    }
    if(cnt&1)
    {
        reverse(b.begin(),b.end());
        reverse(s1.begin(),s1.end());
        s= b+ s+a;
    }
    else
    {
        reverse(a.begin(),a.end());
        s= a+s+b;
    }
    cout<<s<<endl;
}