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>
using namespace std;
using ll = long long;
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define FORD(i,a,b) for(int i=a;i>=b;i--)
#define ALL(a)  (a).begin(),(a).end()
#define SZ(a) int((a).size())
#define SORT(c) sort((c).begin(),(c).end())
#define MEMSET(v, h) memset((v), h, sizeof(v))
#define sqr(x) ((x) * (x))
#define bit(n) (1LL<<(n))
#define pcnt(x) __builtin_popcountll(x)
#define PB push_back
#define MP make_pair
#define endl "\n"
using Edge = pair<int,int>;
using Graph = vector<vector<Edge>>;
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; }
// ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
// ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

int main() {
  string s; cin >> s;
  int q; cin >> q;
  bool isrev = false;
  FOR(i,0,q-1){
    int t; cin >> t;
    if(t==1){
      isrev = !isrev;
    }
    else{
      int f; cin >> f;
      string c; cin >> c;
      if(isrev) f = (f+1);
      if(f==3) f=1;

      if(f==1){
        s = c + s;
      } else {
        s += c;
      }
    }
  }
  if(isrev){
    string copy = s;
    reverse(copy.begin(),copy.end());
    s = copy;
  }
  cout << s << endl;
  
}