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;
typedef long long ll;
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define INF (1LL<<50)
#define MOD 1000000007

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    string S;
    S.reserve(1000000);
    cin >> S;
    int Q;
    cin >> Q;
    bool flg = true; //Tなら文字の先頭がはじめ.
    string ans = S;
    string mae="",ato="";
    mae.reserve(200005);
    ato.reserve(200005);
    int q;
    int f;
    string c;
    rep(i,Q){
        cin >> q;
        if(q==1){
            flg = !flg;
        }else{
            cin >> f;
            cin >> c;
            if(f==1){
                if(flg){
                    mae+=c;
                }else{
                    ato+=c;
                }
            }else{
                if(flg){
                    ato+=c;
                }else{
                    mae+=c;
                }
            }
        }
    }
    if(flg){
        ans+=ato;
        reverse(mae.begin(),mae.end());
        ans = mae+ans;
        cout << ans << endl;
        return 0;
    }
    ans += ato;
    reverse(ans.begin(),ans.end());
    ans+=mae;
    cout << ans << endl;
    return 0;
}