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 rep(i, a, b) for (int i = a; i < (int)(b); i++)
#define rrep(i, a, b) for (int i = b - 1; i >= int(a); i--)

using namespace std;

using ll = long long;
using vi = vector<int>;

bool exor(bool a, bool b) {
    if ((a && b) or (!a && !b)) {
        return false;
    } else {
        return true;
    }
}

int main() {
    string tmp;
    int Q;
    cin >> tmp >> Q;

    deque<char> S;
    rep(i, 0, tmp.size()) { S.push_back(tmp[i]); }

    bool forward = true;
    rep(i, 0, Q) {
        int t;
        cin >> t;

        if (t == 1) {
            forward = !forward;
        } else {
            int f;
            char c;
            cin >> f >> c;

            if (exor(f == 1, forward)) {
                S.push_back(c);
            } else {
                S.push_front(c);
            }
        }
    }

    if (forward) {
        while (!S.empty()) {
            cout << S.front();
            S.pop_front();
        }
    } else {
        while (!S.empty()) {
            cout << S.back();
            S.pop_back();
        }
    }
    printf("\n");

    return 0;
}