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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i=0; i<(n); ++i)
using ll = long long;
const int INF = 1001001001;
const int mod = 1000000007;

int main(){
    deque<char> S;
    char s;
    cin >> s;
    S.push_back(s);
    int n;
    cin >> n;
    int a=0;
    for(int i=0;i<n;i++){
        int t;
        cin >> t;
        if (t==1){
            a+=1;
            continue;
        }
        if(t==2){
            int f;
            char c;
            cin >> f >> c;
            
            if(a%2 == 0 && f ==2){ 
                S.push_back(c);
            }
            else if (a%2 == 0 && f ==1){
                S.push_front(c);
            }
            else if (a%2 == 1 && f ==1){
                S.push_back(c);
            }
            else if (a%2 == 1 && f ==2){
                S.push_front(c);
            }
        }
    }

    if (a%2 == 0){
        for_each(S.begin(), S.end(), [](char x) {cout << x;});
      	cout<<endl; 
        return 0;
    }
    else if(a%2 ==1){
        reverse_iterator<decltype(S)::iterator> first(S.end());
        reverse_iterator<decltype(S)::iterator> last(S.begin());
        for_each(first, last, [](char x) {cout << x;});
    	cout << endl;
    }
    
}