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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
using P = pair<ll, ll>;
using V = vector<ll>;
#define rep(i, n) for(ll i=0;i<(ll)(n);i++)
#define rep1(i, n) for(ll i=1;i<=(ll)(n);i++)
#define rep2(i, m, n) for(ll i=m;i<(ll)(n);i++)
#define rrep(i, n, m) for(ll i=n;i>=(ll)(m);i--)
#define ALL(obj) (obj).begin(), (obj).end()
#define MOD 1000000007
#define INF 1000000000

void solve_abc() {
	string s;

	cin >> s;

	ll q;
	cin >> q;
	ll t, f;
	char c;
	bool is_reversed = false;
	rep(i, q){
		cin >> t;
		if (t==1){
			is_reversed = !is_reversed;
		}
		else {
			cin >> f >> c;
			if (f==1){
				if (is_reversed){
					s = s + c;
				}
				else{
					s = c + s;
				}
			}
			else {
				if (is_reversed){
					s = c + s;
				}
				else{
					s = s + c;
				}
			}
		}
	}

	if (is_reversed){
		reverse(s.begin(), s.end());
	}

	cout << s << "\n";
}

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

	solve_abc();

	return 0;
}