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>
using intl = long long;
#define rep(i, n) for(intl i = 0; i < (intl)(n); i++)
#define repd(i, n) for(intl i = (intl)((n) - 1); i >= 0; i--)
#define repi(i, a, b) for(intl i = (intl)(a); i < (intl)(b); i++)
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
#define INF = 1e9;
#define debug(x) cout << #x << ":" << x << endl
using namespace std;
void fp(bool flag){
    if(flag)cout << "Yes" << endl;
    else cout << "No" << endl;
}
int main(){
    string s;
    cin >> s;
    deque<char> d;
    rep(i,s.length()){
        d.push_back(s[i]);
    }
    intl q;
    cin >> q;
    rep(i,q){
        intl a;
        cin >> a;
        if(a == 1){
            reverse(d.begin(),d.end());
        }
        else if(a == 2){
            intl b;
            cin >> b;
            char t;
            cin >> t;
            if(b == 1){
                d.push_front(t);
            }
            else if(b == 2){
                d.push_back(t);
            }
        }
    }
    rep(i,d.size()){
        cout << d[i];
    }
    return 0;
}