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)

// cities to vapori-
// corner cases // int vs ll // cin vs scanf // clear structures // statement // doublesz // math is your friend
#include <bits/stdc++.h>
#define pb push_back
#define fst first
#define snd second
#define sz(x) int(x.size())
#define fill(x,v) memset(x,v,sizeof(x))
#define REP(i,a,b) for(int i = int(a); i < int(b); ++i)
#define trace(x) cout << #x << " = " << x << endl
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
using namespace std;
typedef long long ll;
typedef pair<int,int> ii;

int main(){
	
	fastio;
	deque<char> dq;
	string s;
	cin >> s;
	for(char ch : s){
		dq.push_back(ch);
	}

	int rev = 0,q,t,f;
	char ch;
	cin >> q;
	while(q--){
		cin >> t;
		if(t == 1){
			rev ^= 1;
		}else{
			cin >> f >> ch;
			if(rev){
				if(f == 1) dq.push_back(ch);
				else dq.push_front(ch);
			}else{
				if(f == 1) dq.push_front(ch);
				else dq.push_back(ch);
			}
		}
	}

	string ans;
	if(rev){
		while(sz(dq)){
			ans += dq.back();
			dq.pop_back();
		}
	}else{
		while(sz(dq)){
			ans += dq.front();
			dq.pop_front();
		}
	}

	cout << ans << endl;

	return 0;
}