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<cstdio>
#include<cmath>
#include<climits>
#include<algorithm>
#include<sstream>
#include<iostream>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<list>
#include<map>
#include<vector>

using namespace std;


#define rep(i,n) for(int i=0; i<n; i++)
#define rept(i,a,b) for(int i=a; i<b; i++)
#define ll long long
#define MOD 1000000007

#define IMIN INT_MIN;
#define IMAX INT_MAX;

string reverseStr(string& str)
{
    int n = str.length();

    // Swap character starting from two
    // corners
    for (int i = 0; i < n / 2; i++)
        swap(str[i], str[n - i - 1]);
    return str;
}
int main(){
    string S;
    int Q;
    cin >> S >> Q;
    int T, F;
    string C;
    string beg="", end="";
    // false is normal
    bool flag = false, flag2;
    rep(i, Q) {
        cin >> T;
        if (T == 1)
            flag = !flag;
        else {
            cin >> F >> C;
            flag2 = ((!flag&& F == 1) || (flag&& F == 2));
            if (flag2) beg += C;
            else end += C;
        }
    }
    if (!flag) {
        if (beg != "") S = reverseStr(beg) + S;
        if (end != "") S = S + end;
    }
    else {
        if (end != "") S = reverseStr(end) + reverseStr(S);
        if (beg != "") S = S + beg;
    }
    cout << S << endl;



	return 0;
}