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 <iostream>
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <math.h>
#include <numeric>
#include <iomanip>
#include <map>

#define REP(i, n ,N) for(ll i = (n); i < (N); i++)
//#define RREP(i, n ,N) for(ll i = (N-1); i >= (n); i--)
#define p(s) cout<<(s)<<endl
#define p2(a, b) cout<<(a)<<" "<<(b)<<endl

using namespace std;
typedef long long  ll;

const long long MOD = 1e9+7;

string JoinWord(string s, string c, int f, bool flag){
    if(flag){
        if(f == 1){return c + s;}
        else{return s + c;}
    } else {
        if(f == 1){return s + c;}
        else{return c + s;}
    }
}

int main(){

    string s;
    int n;
    cin >> s >> n;

    bool reverse_flag = true;

    int t, f;
    string c;
    REP(i, 0, n){
        cin >> t;
        if(t == 1){ reverse_flag = not reverse_flag;}
        else{
            cin >> f >> c;
            s = JoinWord(s, c, f, reverse_flag);
        }

    }

    if(not reverse_flag){reverse(s.begin(), s.end());}

    p(s);

    return 0;
}