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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define rep(i,n) for(int i = 0;i<n;i++)
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0;}
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0;}


int main() {
    string s;
    cin >> s;
    int q;
    cin >> q;
    int cnt = 0;
    string c;
    rep(i,q) {
        int t;
        cin >> t;
        if(t==1) {
            cnt++;
        }
        if(t==2) {
            int f;
            cin >> f;
            cin >> c;
            if(f==1) {
    
                if(cnt%2==0) {
                    c += s;
                } else {
                    s += c;
                }
            }
            if(f==2) {
    
                if(cnt%2==0) {
                    s += c;
                } else {
                    c += s;
                }
            }
        }
    }
    if(cnt%2==1) {
        if(s > c) reverse(s.begin(),s.end());
        else reverse(c.begin(),c.end());
    }
    cout << s << endl;
}