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++ (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

#pragma warning(disable: 4996)
#include <string>
#include <vector>
#include <iostream>
#include <cstdio>
#include <sstream>
#include <fstream>
#include <math.h>
#include <algorithm>

using namespace std;
typedef long long ll;
#define rep(i,n) for (int i = 0; i < (n); i++)

int main() {
    string s;
    int q, t;
    cin >> s >> q;
    int chk = 0;
    rep(i, q) {
        cin >> t;
        if (t == 1)chk++;//chkが偶数の時は入れ替わりなし、奇数では入れ替わっている状態
        if (t == 2) {
            int f;
            string c;
            cin >> f >> c;
            if (f == 1) {
                if (chk % 2 == 0) {//そのまま前
                    s = c + s;
                }
                else {//入れ替わり状態なので後ろに
                    s = s + c;
                }
            }
            else {
                if (chk % 2 == 0) {//そのまま前
                    s = s + c;
                }
                else {//入れ替わり状態なので後ろに
                    s = c + s;
                }
            }
        }
    }
    int n = s.size();
    if (chk % 2 == 0) cout << s << endl;
    else {
        rep(i,n) {
            cout << s[n - 1 - i];
        }
    }
    return 0;
}