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 <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <random>
#include <bitset>
#include <queue>
#include <cmath>
#include <deque>
#include <unordered_map>
#include <set>
#include <map>
#define rep(i,n) for (int i=0; i<n;++i)
#define ALL(a)  (a).begin(),(a).end()
#define rep_down(i,n) for (int i=n-1; i>=0;--i)
typedef long long ll;
using namespace std;
int Q;
string S;
deque<char> D;
struct edge{bool mae; char c;};

int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);

  cin >> S;
  rep(i, S.size()) {
    D.push_back(S[i]);
  }
  cin >> Q;
  int a,f;
  int cnt = 0;
  char c;
  vector<edge> V;
  rep(i, Q) {
    cin >> a;
    if (a==1) {
      cnt++;
    } else {
      cin >> f >> c;
      f+=cnt%2;
      f%=2;
      V.push_back({f==0, c});
    }
  }
  rep(i, Q-cnt) {
    edge e = V[i];
    if (e.mae) D.push_back(e.c);
    else D.push_front(e.c);
  }
  cnt%=2;
  if (cnt>0) {
    while(!D.empty()) {
      char d = D.back(); D.pop_back();
      cout << d;
    }
  }else {
    while(!D.empty()) {
      char d = D.front(); D.pop_front();
      cout << d;
    }
  }
  cout << "\n";
  return 0;
}