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>

#define SORT(v) sort(v.begin(),v.end())
#define si(n) scanf("%d",&n)
#define sii(n,m) scanf("%d %d",&n,&m)
#define sl(n) scanf("%lld",&n)
#define sll(n,m) scanf("%lld %lld",&n,&m)
#define ss(cad) scanf("%s",cad)
#define PB push_back
#define fst first
#define scn second
#define DBG(x) cerr << #x << " = " << (x) << endl; 
#define M 1000000007
#define N_MAX 100010

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<ll> vl;
typedef pair<int,int> pi;
typedef vector<pi> vp;

char cad[N_MAX];

int main()
{
	ss(cad);
	int k;
	si(k);
	vector<char> ini, fin;
	int curr = 1; // String original
	while(k--)
	{
		int t, f;
		char c;
		si(t);
		if(t == 1)
			curr *= (-1);
		else
		{
			si(f);
			scanf(" %c", &c);
			if(f == 1)
			{
				if(curr == 1)
					ini.push_back(c);
				else
					fin.push_back(c);
			}
			else
			{
				if(curr == 1)
					fin.push_back(c);
				else
					ini.push_back(c);
			}
		}
	}
	int n = strlen(cad);
	if(curr == 1)
	{
		for(int i = ini.size() - 1; i >= 0; i--)
			printf("%c", ini[i]);
		for(int i = 0; i < n; i++)
			printf("%c", cad[i]);
		for(int i = 0; i < fin.size(); i++)
			printf("%c", fin[i]);
		printf("\n");
	}
	else
	{
		for(int i = fin.size() - 1; i >= 0; i--)
			printf("%c", fin[i]);
		for(int i = n-1; i >= 0; i--)
			printf("%c", cad[i]);
		for(int i = 0; i < ini.size(); i++)
			printf("%c", ini[i]);
		printf("\n");

	}
    return 0;
}