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: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <deque>

using namespace std;

#define rep1(i, n) for (long long i = 0; i < (long long)(n); i++)
#define rep2(i, n, m) for (long long i = (long long)(n); i <= (long long)(m); i++)
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a > b) { a = b; return 1; } return 0; }
const long long mod = 1e9 + 7;

int main() {
    string s; cin >> s;
    long long q; cin >> q;
    deque<string> queue;
    bool rev = false;
    queue.push_back(s);

    rep1(i, q) {
        int t; cin >> t;
        if (t == 1) {
            rev = !rev;
        } else {
            int f; cin >> f;
            string c; cin >> c;
            if (rev) {
                if (f == 1) {
                    queue.push_back(c);
                } else {
                    queue.push_front(c);
                }
            } else {
                if (f == 1) {
                    queue.push_front(c);
                } else {
                    queue.push_back(c);
                }
            }
        }
    }
    if (rev) {
        while (!queue.empty()) {
            cout << queue.back(); queue.pop_back();
        }
    } else {
        while (!queue.empty()) {
            cout << queue.front(); queue.pop_front();
        }
    }
    cout << "" << endl;
}