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 <iostream>
#include <string>
#include <map>
#include <queue>
#include <math.h>
#include <bitset>
#include <cmath>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
#define rep(i,s,n)for(int i = s;i<n;i++)
#define repe(i,s,n)for(ll i = s;i<=n;i++)
#define rep_r(i,s,n)for(ll i = n-1;i>s;i--)
#define rep_re(i,s,n)for(ll i = n-1;i>=s;i--)
ll MOD = 1e9 + 7;

char C[500001];
int main() {
	string s; cin >> s;
	rep(i, 0, s.size()) {
		C[200000 + i] = s[i];
	}
	ll C_Pr = 199999;
	ll C_Po = 200000 + s.size();
	
	ll q; cin >> q;
	bool o = true;
	rep(i, 0, q) {
		ll Q; cin >> Q;
		if (Q == 1) {
			o = !o;
		}
		else {
			ll f; char c; cin >> f >> c;

			if (o && f == 1 || !o && f == 2) {
				C[C_Pr--] = c;
			}
			else if (o && f == 2 || !o && f == 1) {
				C[C_Po++] = c;
			}
		}
	}
	if (o) {
		rep(i, C_Pr+1, C_Po) {
			cout << C[i];
		}
	}
	else {
		rep_r(i, C_Pr, C_Po) {
			cout << C[i];
		}
	}
	cout << endl;

	return 0;
}