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)

#pragma comment(linker, "/STACK: 2000000")
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define clock (clock() * 1000.0 / CLOCKS_PER_SEC)
#define dbg(x) cout << #x << '=' << x << '\n';
#define ll long long
#define double long double
#define x first
#define y second
#define pi pair <int, int>
#define vi vector <int>
#define L nod<<1
#define R ((nod<<1)|1)
#define int ll
const ll mod = 998244353;
string s;
int q, rev;

int32_t main(){
	ios_base :: sync_with_stdio(0); cin.tie(); cout.tie();
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	cin>>s>>q;
	while(q--){
		int t;
		cin>>t;
		if(t==1) rev++, rev%=2;
		else{
			int f; char c;
			cin>>f>>c;
			if((rev+f)%2==1) s=c+s;
			else s=s+c;
		}
	}
	if(rev) reverse(s.begin(), s.end());
	cout<<s;
}