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<iostream>
#include<stdio.h>
//#include <bits/stdc++.h>
#include<vector>
#include<float.h>
#include<iomanip>
#include<algorithm>
#include<string>
#include<cstring>
#include<math.h>
#include<cmath>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include <cassert>
#include <cmath>
#include<cstdint>

#define INF 1e9
#define rep(i,n)for(int i=0;(i)<(int)(n);i++)
#define REP(i,a,b)for(int i=(int)(a);(i)<=(int)(b);i++)
#define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
#define vec(type,n) vector<type>(n)
#define vvec(m,n) vector<vector<int>> (int(m),vector<int>(n))
#define ALL(a)  (a).begin(),(a).end()

using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
using P = pair<int,int>;

int main(){
    string s;
    int q;cin>>s>>q;
    int cnt = 0;
    int d = 0;
    rep(i,q){
        int a;cin>>a;
        if(a == 1)cnt++;
        if(a == 2){
            int b;
            char c;cin>>b>>c;
            if(cnt%2==1){
                d++;
                if(b == 1)b = 2;
                else if(b == 2)b = 1;
            }
            if(b == 1)s.insert(s.begin(),c);
            if(b == 2)s.push_back(c);
        }
    }
    if(cnt%2==1)reverse(ALL(s));
    cout<<s<<endl;
}