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>

#define rep(i, a, n)  for(ll i=a; i<n; i++)
#define per(i, a, n)  for(ll i=n-1; i>=a; i--)
#define fill0(n)  setfill('0') << right << setw(n)
#define all(x)  (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define LONGMAX 1e18
#define INTMAX 1000000000

using namespace std;
typedef long long ll;
typedef pair<ll ,ll> P;
const ll mod = 1e9+7;
const long double EPS = 0.0000000001;

string s;
int q;
int main(){
	cin >> s;
	cin >> q;
	vector<char> v(all(s));
	rep(i,0,q){
		int t;
		cin >> t;
		if(t == 1)reverse(all(v));
		else{
			int f;
			char c;
			cin >> f >> c;
			if(f == 1)v.insert(v.begin(), c);
			else v.push_back(c);
		}
	}
	for(int i = 0; i < v.size(); i++)cout << v[i];
	cout << endl;

}