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;

typedef long long ll;
typedef pair<ll, ll> P;

const ll MOD = 1e9 + 7;
const ll inf = 1LL<<60;
const double eps = 1e-9;

const int dx[4] = { 1, 0, -1, 0 };
const int dy[4] = { 0, 1, 0, -1 };
 
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main()
{
    string s;
    cin >> s;
    int q;
    cin >> q;
    int x, f;

    string bef, aft;
    char c;
    int ok = 1;

    for (int i = 0; i < q; i++) {
        cin >> x;
        if (x == 1) ok = 1 - ok;
        if (x == 2) {
            cin >> f >> c;
            if (ok) {
                if (f == 1) bef = c + bef;
                if (f == 2) aft = aft + c;
            } else {
                if (f == 1) aft = aft + c;
                if (f == 2) bef = c + bef;
            }
        }
    }
    string tmp = bef + s + aft; 
    if (ok) cout << tmp << endl;
    else {
        reverse(tmp.begin(), tmp.end());
        cout << tmp << endl;
    }
    return 0;
}