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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include<limits.h>
#include<iomanip>
#include <cassert>
#include <list>
#define rep(i, n) for(int i = 0; i < n; i++)
#define rrep(i, n) for(int i = n-1; i >= 0; i--)
#define REP(i, m, n) for(int i = m; i < n; i++)
#define ALL(v) v.begin(),v.end()
#define RALL(v) v.rbegin(),v.rend()
#define Cout(x) cout << x << endl;
#define check(v) rep(i,v.size()) cout << v[i] << endl;
#define vint(x) vector<int> x
#define vll(x) vector<long long> x
#define vstr(x) vector<string> x
const long long LLINF = 1LL<<60;
const int INF = 1e9;
const long long MOD = 1000000007;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
typedef long long ll;
#define pint pair<int,int>
#define pll pair<ll,ll>
using namespace std;
//オーバーフローに気をつけろよおおおおおお
//確認忘れるなよおおおおおお

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string s; cin >> s;
    int q; cin >> q;
    list<char> V;
    rep(i,s.size()) V.push_back(s[i]);

    int flag = 1;
    rep(i,q) {
        int t; cin >> t;
        if(t==2) {
            int f; cin >> f;
            char c; cin >> c;
            if(flag) {
                if(f==1) V.insert(V.begin(),c);
                else V.push_back(c);
            }
            else {
                if(f==2) V.insert(V.begin(),c);
                else V.push_back(c);
            }
        }
        else {
            flag ^= 1;
        }
    }

    if(!flag) {
        reverse(ALL(V));
    }
    for(auto val : V) cout << val;
    cout << endl;

    return 0;
}