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 rep(i,n) for(int i = 0;i < n;++i)
#define rep2(i,n,m) for(int i = n;i < m;++i)
#define all(v) v.begin(),v.end()

using namespace std;
using ll = long long;
const int INF = 1001001001;

uintmax_t combination(unsigned int n, unsigned int r) {
  if ( r * 2 > n ) r = n - r;
  uintmax_t dividend = 1;
  uintmax_t divisor  = 1;
  for ( unsigned int i = 1; i <= r; ++i ) {
    dividend *= (n-i+1);
    divisor  *= i;
  }
  return dividend / divisor;
}

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

    int n;
    string s;
    deque<string> v;
    cin >> s >> n;

    v.push_back(s);
    rep(i,n){
        int a;
        cin >> a;
        if(a == 1){
            reverse(all(v));
        }else if(a == 2){
            int b;
            string st;
            cin >> b >> st;
            if(b == 1){
                v.push_front(st);
            }else if(b == 2){
                v.push_back(st);
            }

        }
    }
    rep(i,v.size()){
        cout << v.at(i);
    }
    cout << endl;
    return 0;
}