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 <algorithm>     //sort,二分探索,など
#include <bitset>        //固定長bit集合
#include <cmath>         //pow,logなど
#include <complex>       //複素数
#include <deque>         //両端アクセスのキュー
#include <functional>    //sortのgreater
#include <iomanip>       //setprecision(浮動小数点の出力の誤差)
#include <iostream>      //入出力
#include <iterator>      //集合演算(積集合,和集合,差集合など)
#include <map>           //map(辞書)
#include <numeric>       //iota(整数列の生成),gcdとlcm(c++17)
#include <queue>         //キュー
#include <set>           //集合
#include <stack>         //スタック
#include <string>        //文字列
#include <unordered_map> //イテレータあるけど順序保持しないmap
#include <unordered_set> //イテレータあるけど順序保持しないset
#include <utility>       //pair
#include <vector>        //可変長配列

using namespace std;
typedef long long ll;

//マクロ
#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define REPD(i, n) for (ll i = (ll)(n)-1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = (a); i <= (b); i++)
#define FORD(i, a, b) for (ll i = (a); i >= (b); i--)
#define ALL(x) (x).begin(), (x).end() //sortなどの引数を省略したい
#define SIZE(x) ((ll)(x).size())      //sizeをsize_tからllに直しておく
#define MAX(x) *max_element(ALL(x))
#define INF 1000000000000 //10^12
#define MOD 10000007      //10^9+7
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define MAXR 100000 //10^5:最大のrange(素数列挙などで使用)

signed main()
{
  ll Q;
  string S;

  bool is_reversed = false;

  cin >> S;
  cin >> Q;

  REP(i, Q)
  {
    ll t;
    cin >> t;
    if (t == 1)
    {
      is_reversed = !is_reversed;
    }
    else if (t == 2)
    {
      ll f;
      string s;
      cin >> f >> s;
      if (f == 1)
      {
        if (!is_reversed)
        {
          S = s + S;
        }
        else
        {
          S = S + s;
        }
      }
      else if (t == 2)
      {
        if (!is_reversed)
        {
          S = S + s;
        }
        else
        {
          S = s + S;
        }
      }
    }
  }

  if (is_reversed)
  {
    reverse(ALL(S));
  }
  cout << S << endl;
}