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 mk make_pair
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef long double ld;
int a[100010];
int main(){
    int n, q;
    string s;
    while(cin>>s>>q){
    	n=s.size();
        ll ans=0;
        int x, f;
        bool l=0;
        char c[111];
        deque<char> v;
        for(int i=0; i<n; ++i){
        	v.push_back(s[i]);
        }
        for(int i=0; i<q; ++i){
            scanf("%d",&x);
            if(x==1){
            	l=!l;
            }
            else{
            	scanf("%d%s",&f,c);
            	if(l){
            		if(f==1){
            			v.push_back(c[0]);
            		}
            		else{
            			v.push_front(c[0]);
            		}
            	}
            	else{
            		if(f==2){
            			v.push_back(c[0]);
            		}
            		else{
            			v.push_front(c[0]);
            		}
            	}
            }
        }
        if(l){
        	reverse(v.begin(), v.end());
        }
        for(auto it:v){
        	printf("%c",it);
        }
        cout<<endl;
    }
  return 0;
}