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)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;

#define rep( i, n, m ) for(int i = ( n ); i < ( m ); i++)
#define rep_d( i, n, m ) for(int i = ( n ) - 1; i >= ( m ); i--)
#define sort_asc( X ) sort(( X ).begin(), ( X ).end())
#define sort_desc( X ) sort(( X ).begin(), ( X ).end(), greater <>())
#define endl "\n"

template <class T> bool chmax( T& a, T b );
template <class T> bool chmin( T& a, T b );
int* eratosthenes( int N );
bool bit_search( int pattern, int N );
void initialize( void );

const ll INF = 1LL << 60;
const int k_mod = 1e9 + 7;
template <typename T, typename K> struct P {
	T a; K b;
};
int main( void ){
	// initialize();
	string S, L, R;
	cin >> S;
	int Q;
	cin >> Q;
	bool is_inv = false;
	rep( reppp, 0, Q ){
		int input; cin >> input;
		if (input == 1){
			is_inv = !is_inv;
		}
		else {
			int in; string c;
			cin >> in >> c;
			if (in == 1){
				if (!is_inv){
					L.insert(0, c );
				}
				else {
					R.append( c );
				}
			}
			else {
				if (!is_inv)
					R.append( c );
				else
					L.insert(0, c );

			}
		}
	}
	if (!is_inv)
		printf( "%s%s%s\n", L.c_str(), S.c_str(), R.c_str());
	else {
		int K = R.size();
		rep_d( i, K, 0 )
		printf( "%c", R[ i ] );
		K = S.size();
		rep_d( i, K, 0 )
		printf( "%c", S[ i ] );
		K = L.size();
		rep_d( i, K, 0 )
		printf( "%c", L[ i ] );
		cout << endl;
	}
}

template <class T> bool chmax( T& a, T b ) {
	if (a < b){
		a = b; return true;
	}
	return false;
}
template <class T> bool chmin( T& a, T b ) {
	if (a > b){
		a = b; return true;
	}
	return false;
}
int* eratosthenes( int N ){
	int* prime_array = new int[ N + 1 ];
	int lim = N;
	prime_array[ 0 ] = 0;
	prime_array[ 1 ] = 0;
	rep( i, 2, N + 1 ){
		prime_array[ i ] = 1;
	}
	rep( i, 2, lim ){
		if (prime_array[ i ] == 0)
			continue;
		lim = N / i;
		for(int j = i * 2; j < N + 1; j += i){
			prime_array[ j ] = 0;
		}
	}
	return prime_array;
}
bool bit_search( int pattern, int N ){
	int cnt = 0;
	rep( bit, 0, N ){
		if (pattern & ( 1 << bit )){
			cnt++;
		}
	}
	return true;
}
void initialize( void ){
	cout << fixed << setprecision( 10 );
}