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)

    #pragma GCC optimize("O3")
    #pragma comment(linker, "/stack:200000000")
    #pragma GCC optimize("unroll-loops")
     
    #include <algorithm>
    #include <bitset>
    #include <cassert>
    #include <chrono>
    #include <cmath>
    #include <complex>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <ctime>
    #include <deque>
    #include <functional>
    #include <iomanip>
    #include <iostream>
    #include <iterator>
    #include <limits>
    #include <climits>
    #include <list>
    #include <map>
    #include <numeric>
    #include <queue>
    #include <random>
    #include <ratio>
    #include <set>
    #include <sstream>
    #include <stack>
    #include <string>
    #include <unordered_map>
    #include <unordered_set>
    #include <utility>
    #include <vector>
    using namespace std;
     
    typedef long long ll;
    typedef long double LD;
    #define int ll
    #define double LD
    #define pb push_back
    #define mp make_pair
    #define REP(i,n) for (int i = 0; i < n; i++)
    #define loop(i,n) for (int i = 0; i < n; i++)
    #define FOR(i,a,b) for (int i = a; i < b; i++)
    #define REPD(i,n) for (int i = n-1; i >= 0; i--)
    #define FORD(i,a,b) for (int i = a; i >= b; i--)
    #define remax(a,b) a = max(a,b)
    #define remin(a,b) a = min(a,b)
    #define mem(a, b) memset(a, b, sizeof a)
    #define append push_back
    #define all(v) v.begin(),v.end()
    #define F first
    #define S second
    #define print(x) cout << (x) << endl
    #define PRECISION(x) cout << fixed << setprecision(x)
    #define FAST_IO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
     
    const double PI=acos(-1);
    const int MOD = 1000000007;
    const int FMOD = 998244353;
    const double eps = 1e-9;
     
    mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count()); 
  
  
	signed main()
	{
		string s;
		cin>>s;
		int q;
		cin>>q;
		while(q--)
		{
			int t;
			cin>>t;
			if(t==1)
			{
				reverse(s.begin(),s.end());
			}
			else
			{
				int r;
				char c;
				cin>>r>>c;
				if(r==1)
				{
					s=c+s;
				}
				else
				{
					s=s+c;
				}
			}
			
		}
		cout<<s;
	}