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"

using namespace std;

using ll = long long;

template<typename T> using V = std::vector<T>;
using Vi = V<int>;
using VVi = V<V<int>>;
using Vl = V<ll>;
using VVl = V<V<ll>>;
using Vs = V<string>;

template<typename T1, typename T2> using P = std::pair<T1, T2>;
using Pii = P<int, int>;
using Pll = P<ll, ll>;
using Pdd = P<double, double>;

template<typename T1, typename T2> using M = std::map<T1, T2>;
using Mii = M<int, int>;
using Msi = M<string, int>;

#define REP(i,n) for(int i=0;i<(int)(n);++i)
#define REP2(i,s,e) for(int i=(int)(s);i<(int)(e);++i)
#define RREP(i,s,e) for(int i=(int)(s);i>=(int)(e);--i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin(),ie=(c).end();i!=ie;++i)
#define ALL(c) (c).begin(), (c).end()

const double PI = acos(-1);
const ll MOD = 1e9+7;

string s;
int q;

int main() {
    cin >> s >> q;

    deque<char> dq;
    for (auto c: s) {
        dq.push_back(c);
    }

    bool f = true;
    REP(i, q) {
        int t; cin >> t;
        if (t == 1) {
            f = !f;
        } else {
            int x; char c; cin >> x >> c;
            if ((x == 1 && f) || (x == 2 && !f)) {
                dq.push_front(c);
            } else {
                dq.push_back(c);
            }
        }
    }

    if (f) {
        for (auto c: dq) {
            cout << c;
        }
    } else {
        RREP(i, dq.size()-1, 0) {
            cout << dq.back();
            dq.pop_back();
        }
    }

    cout << '\n';

    return 0;
}