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>
#include <iostream>
#include <map>
#include <vector>
#include <cmath>
#include <ios>
#include <iomanip>
#include <algorithm>
#define REP(i, n) for (int i = 0; i < (int)(n); ++i)
#define REP2(i, m, n) for (int i = (m); i < (int)(n); ++i)
#define REPR(i, n) for (int i = (n)-1; i >= 0; --i)
#define REPR2(i, m, n) for (int i = (n)-1; i >= (m); --i)
#define REPx(x, a) for(auto x : a)
#define ALL(a) a.begin(), a.end()
#define SORT(a) sort(ALL(a))
#define SORTR(a) sort(ALL(a), greater<int>())
#define REVERSE(a) reverse(ALL(a))
#define bit_search(bit, n) REP(bit, 1<<(n))
#define bit_check(bit, i) (bit>>(i)) & 1
#define setpre(n) fixed << setprecision((n))
#define UNIQUE(a) do {SORT(a); (a).erase(unique(ALL(a)), (a).end());} while(0)
#define MAX(a) *max_element(ALL(a))
#define MIN(a) *min_element(ALL(a))
#define bisect_left(a, x) lower_bound(ALL(a), (x)) - a.begin()
#define bisect_right(a, x) upper_bound(ALL(a), (x)) - a.begin()
#define INPUT(a) REP(i, a.size()) cin >> a[i];
#define INPUTP(a) REP(i, a.size()) cin >> a[i].first >> a[i].second;
#define OUTPUT_PERMUTATION(n) do{VI v(n); iota(ALL(v), 1); do{REPx(x, v) cout << x << " "; ENDL} while(next_permutation(ALL(v)));} while(0);
#define MAKE_PERMUTATION(n) do{VVI a(fact(n), VI(n)); int idx = 0; VI v(n); iota(ALL(v), 1); do{REP(i, n) a[idx][i] = v[i]; idx++; PER = a;} while(next_permutation(ALL(v)));} while(0); // int fact(), VVI PERを宣言しておく
#define ENDL cout << endl;
using namespace std;
using LL = long long;
using LD = long double;
using PII = pair<int, int>;
using VPII = vector<PII>;
using PLL = pair<LL, LL>;
using VPLL = vector<PLL>;
using VI = vector<int>;
using VVI = vector<VI>;
using VLL = vector<LL>;
using VVLL = vector<VLL>;
using VC = vector<char>;
using VS = vector<string>;
using VB = vector<bool>;
using VD = vector<double>;
const int INF = 1e9;
const LL MOD = 1e9 + 7;

template<class T> inline bool chmin(T &a, T b){if (a > b) {a = b; return true;} return false;}
template<class T> inline bool chmax(T &a, T b){if (a < b) {a = b; return true;} return false;}

void END(){cout << 1 << endl; exit(0);}

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  deque<char> dq;
  string s;
  int q;
  cin >> s >> q;
  REPx(x, s) dq.push_back(x);
  bool flg = false;
  REP(i, q)
  {
    int t;
    cin >> t;
    if (t == 1)
    {
      if (flg) flg = false;
      else flg = true;
    }
    else
    {
      int f;
      char c;
      cin >> f >> c;
      if (flg)
      {
        if (f == 1) dq.push_back(c);
        else dq.push_front(c);
      }
      else
      {
        if (f == 1) dq.push_front(c);
        else dq.push_back(c);
      }
    }
  }
  if (flg) REVERSE(dq);
  REPx(x, dq) cout << x;

  return 0;
}