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>
using namespace std;

template<typename T>
istream& operator >> (istream& istr, vector<T>& vec){
    for(T& x: vec) istr >> x;
    return istr;
}

template<typename T>
ostream& operator << (ostream& ostr, vector<T>& vec){
    if(!vec.empty()){
        ostr << vec.front();
        for(auto itr = ++vec.begin(); itr != vec.end(); itr++)
            ostr << "" << *itr;
    }
    return ostr;
}

template<typename T>
ostream& operator << (ostream& ostr, vector<vector<T>>& mat){
    if(!mat.empty()){
        ostr << mat.front();
        for(auto itr = ++mat.begin(); itr != mat.end(); itr++)
            ostr << endl << *itr;
    }
    return ostr;
}

template<typename T, typename U>
ostream& operator << (ostream& ostr, pair<T,U>& p){
    ostr << p.first << " " << p.second;
    return ostr;
}

template<typename T, typename U>
ostream& operator << (ostream& ostr, vector<pair<T,U>>& mat){
    if(!mat.empty()){
        ostr << mat.front();
        for(auto itr = ++mat.begin(); itr != mat.end(); itr++)
            ostr << endl << *itr;
    }
    return ostr;
}

typedef long long ll;
typedef vector<ll> vll;
typedef vector<char> vc;
// vll x(n); x[i]=a;
typedef vector<vector<ll>> vvll;
typedef vector<vector<char>> vvc;
// vvll x(m,vi(n)); x[i][j]=a;
typedef pair<ll,ll> pll;
typedef vector<pair<ll,ll>> vpll;
// vpll x(n); x[i]={a,b}; x[i].first/second;

#define PI 3.1415926535897932
#define INF 1e+12
#define EPS 1e-12
#define REP(i,n) for(ll i=0; i<(n); i++)
#define RREP(i,n) for(ll i=(n)-1; i>=0; i--)
#define FOR(i,a,b) for(ll i=(a); i<(b); i++)
#define FORR(i,a,b) for(ll i=(b)-1; i>=(a); i--)
#define dump(x) cout << #x << " = " << (x) << endl
#define all(x) (x).begin(),(x).end()
// sort(all(x));

int main(){
    string S;
    vc s, r, ans;
    ll q, t, f, x=1;
    char c;

    cin >> S;
    REP(i,S.length()) s.emplace_back(S[i]);
    cin >> q;
    REP(i,q){
        cin >> f;
        if(f==1){
            x*=-1;
            continue;
        }
        cin >> t >> c;
        t = t*2-3;
        if(x*t==1) s.emplace_back(c);
        else r.emplace_back(c);
    }

    if(x==1){
        RREP(i,r.size()) ans.emplace_back(r[i]);
        REP(i,s.size()) ans.emplace_back(s[i]);
    }
    else{
        RREP(i,s.size()) ans.emplace_back(s[i]);
        REP(i,r.size()) ans.emplace_back(r[i]);
    }
    cout << ans << endl;
    return 0;
}