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 <bits/stdc++.h>
#include <string>
#include <vector>
/* 変数i を n回繰り返すマクロ */
#define rep(i,n) for (int i = 0;i < (n);i++)
typedef long long ll;
using namespace std;
using Graph = vector<vector<int>>;
/* map_for : mapでfor文を回す */
/* gcd_lcm : 最大公倍数 & 最小公約数 */
int main()
{
    cout << fixed << setprecision(10);//小数点以下10桁出力
    string a;
    cin >> a;
    int q;
    cin >> q;
    int flag = 1;//前か後flag
    rep(i,q){
        int t;
        cin >> t;
        if(t == 1){
            flag *= -1;//flag反転
        }else{
            int f;
            cin >> f;
            string c;
            string tmp;
            cin >> c;
            if(f == 1){//先頭
                if(flag == 1){
                    tmp = c + a;
                }else{
                    tmp = a + c;
                }
            }else{//末尾
                if(flag == 1){
                    tmp = a + c;
                }else{
                    tmp = c + a;
                }
            }
            a = tmp;
        }
    }

    if(flag == -1){
        reverse(a.begin(),a.end());
    }

    cout << a << endl;

    return 0;
}