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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
constexpr int di[] = { 0, 1, 0, -1 }, dj[] = { 1, 0, -1, 0 };
#define rep(i, n) for (int i = 0; i < n; i++)
#define repr(i, n) for (int i = n-1; i >= 0; i--)
#define SORT(v) sort((v).begin(), (v).end())
#define SORTR(v) sort((v).rbegin(), (v).rend())
#define all(v) (v).begin(), (v).end()
#define vmax 10000
constexpr ll inf = 1e18+7;
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/

int main(){
  cout<<fixed<<setprecision(20);
  cin.tie(0);ios::sync_with_stdio(false);
  string s;
  int q;
  cin>>s>>q;
  int t,f;
  int flag=1;
  char c;
  deque<char> ans;
  rep(i,s.size()){
    ans.push_back(s[i]);
  }
  rep(i,q){
    cin>>t;
    if(t==1){
      flag^=1;
    }else{
      cin>>f>>c;
      f--;
      if(flag^f){
        ans.push_front(c);
      }else{
        ans.push_back(c);
      }
    }
  }
  int n=ans.size();
  if(flag){
    rep(i,n){
      cout<<ans[i];
    }
  }else{
    repr(i,n){
      cout<<ans[i];
    }
  }
  cout<<endl;
  return 0;
}