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: Wrong Answer)

#include <bits/stdc++.h>
#define INF INT_MAX
#define ll  long long
#define ull unsigned long long
#define rep(i,n) for(ll i=0; i<n; ++i)
#define FOR(i, s, e) for(ll i=s; i<e; ++i)
#define MOD 1000000007
using namespace std;
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;}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  string S; cin >> S;
  ll Q; cin >> Q;
  int dir = 1;
  string pre, post;
  rep(i, Q) {
    int T; cin >> T;
    if(T == 1) {
      dir *= -1;
    } else {
      int F; char c;
      cin >> F >> c;
      if(F == 1 && dir == 1) S = c + S;
      else if(F == 1 && dir == -1) S = S + c;
      else if(F == 2 && dir == 1) S = S + c;
      else if(F == 2 && dir == -1) S = c + S;
      else cerr << "Error" << endl;
    }
  }
  if(dir == -1) {
    for(ll i=S.size(); i>=0; --i) cout << S[i];
  } else {
    cout << S;
  }
  cout << endl;
}