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: Time Limit Exceeded)

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;

template <class T>
bool INRANGE(T x, T a, T b) { return a <= x && x <= b; }
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define RREP(i, n) for (int i = (n); i >= 0; --i)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define RFOR(i, a, b) for (int i = (a); i >= (b); --i)
#define ALL(v) (v).begin(), (v).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" \
                      << " " << __FILE__ << endl;

typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vi> vvi;
typedef pair<int, int> pii;

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    string s;
    cin >> s;
    int q;
    cin >> q;

    string ans = s;

    bool is_convert = false;
    REP(i, q) {
        int query;
        cin >> query;

        if (query == 1) {
            is_convert = !is_convert;
        } else {
            int f;
            cin >> f;
            string c;
            cin >> c;

            if (f == 1) {
                if (is_convert) ans += c;
                else ans = c + ans;
            } else {
                if (is_convert) ans = c + ans;
                else ans += c;
            }
        }
    }

    if (is_convert)reverse(ALL(ans));
    cout << ans << endl;

    return 0;
}