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;
#define ll long long
#define REP(i,m,n) for(int i=(int)(m); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define RREP(i,m,n) for(int i=(int)(m); i>=(int)(n); i--)
#define rrep(i,n) RREP(i,n-1,0)
#define REPL(i,m,n) for(ll i=(ll)(m); i<(ll)(n); i++)
#define repl(i,n) REPL(i,0,n)
#define all(v) v.begin(), v.end()
const int inf = 1e9+7;
const ll longinf = 1LL<<60;
const ll mod = 1e9+7;
int main() {
  string S;
  cin >> S;
  deque<char> T;
  for(char c: S) T.push_back(c);
  bool dir = true;
  int Q;
  cin >> Q;
  rep(q, Q) {
    int t;
    cin >> t;
    t--;
    if(!t) {
      dir = !dir;
    } else {
      int f;
      char c;
      cin >> f >> c;
      f--;
      if(!f) {
        if(dir) T.push_front(c);
        else T.push_back(c);
      } else {
        if(dir) T.push_back(c);
        else T.push_front(c);
      }
    }
  }
  if(dir) {
    while(!T.empty()) {
      cout << T.front();
      T.pop_front();
    }
  } else {
    while(!T.empty()) {
      cout << T.back();
      T.pop_back();
    }
  }
  cout << endl;
  return 0;
}