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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <algorithm>
#include <utility>
#include <cmath>
using namespace std;
using ll=long long;
using P=pair<int,int>;
#define rep(i,a,b) for(ll i=a;i<ll(b);i++)
#define repr(i,a,b) for(ll i=a;i>=ll(b);i--)
#define endl '\n'
#define ALL(x) x.begin(),x.end()
#define ALLR(x) x.rbegin(),x.rend()
#define INF 1e9
#define DEBUG(x) cout<<"debug: "<<x<<endl
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}

string s;
int q;
int main(){
	cin >> s >> q;
	rep(_, 0, q){
		int t; cin >> t;
		if(t-1){
			int f; char c; cin >> f >> c;
			if(f-1) s += c;
			else s = c + s;
		}
		else{
			int sl = s.length();
			char  top = s[0],  back = s[sl-1];
			s[0] = back;
			s[sl-1] = top;
		}
	}
	cout << s << endl;
    return 0;
}