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>
#define rep(i, n) for(long long int i = 0; i < n; i++)
#define _rep(i, m, n) for(long long int i = m; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
//const int N = 1000000;
const ll mod = 1000000007;
using Graph = vector<vector<int>>;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
  
ll gcd(ll A, ll B){
    if(B == 0)return A;
    return gcd(B, A % B);
}
ll lcm(ll A, ll B){
    return A * (B / gcd(A, B));
}
  
/*------------------------------------------------------------------*/
int main(){
    string s; cin >> s;
    int q; cin >> q;
    int m;
    int f;
    char c;

    while(cin >> m){
        if(m == 1){
            reverse(s.begin(), s.end());
        }else{
            cin >> f;
            cin >> c;
            if(f == 1){
                s = c + s;
            }else{
                s = s + c;
            }
        }
    }
    cout << s << endl;
}