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 ll          long long
#define ld          long double
#define pb          push_back
#define pf          push_front
#define mp          make_pair
#define pll         pair<ll,ll>
#define vl          vector<ll>
#define sl          set<ll> 
#define vll         vector<pll>
#define ml          map<ll,ll>
#define mll         map<pll,ll>
#define all(a)      a.begin(),a.end()
#define x           first
#define y           second
#define sz(x)       (ll)x.size()
#define dl          '\n'
#define mod         (ll)1000000007
#define lp(i,a,b)   for(ll i=a;i<b;++i)
#define lpr(i,a,b)  for(ll i=a;i>=b;i--)
#define lpd(i,x)    for(auto i:x)
#define ios         ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
ll inf = (ll)1e18;
ld pi = 3.141592653589;
using namespace std;

ll bpow(ll x, ll v) {   
    if (v == 0)
        return 1 % mod;
    if (v % 2 == 1)
        return (x * bpow(x, v - 1)) % mod;
    ll r = bpow(x, v / 2);
    return (r * r) % mod;   
}

ll solve(){ 
    string a; cin>>a;
    deque<char> d;
    lp(i,0,sz(a)) d.pb(a[i]);
    ll q,i=0,c=0; cin>>q;
    while(q--){
        ll p; cin>>p;
        if(p == 2){
            ll f; cin>>f;
            char g; cin>>g;
            if(c%2 and f==1) d.pb(g);
            else if(c%2) d.pf(g);
            if(c%2==0 and f==1) d.pf(g);
            else if(c%2==0) d.pb(g); 
        }
        else c++;
    }
    if(c%2)
    {
        for(auto i = d.rbegin();i!=d.rend();i++)
            cout<<*i;
    }
    else {
        lpd(i,d) cout<<i;
    }
}

int main(){
    ios
    ll t=1; 
    //cin>>t;
    while(t--){
        solve();
    }
}