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)

/* 実行時間1s制限の場合10^6 1,000,000 : 余裕をもって間に合う10^7
  10,000,000 : おそらく間に合う10^8 100,000,000 : 非常にシンプルな処理
  でないと厳しい
 */
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)
#include <bits/stdc++.h>
using namespace std;

typedef long long ll; // long longはデータモデル(LLP64, LP64, ...)によらず64bit.
ll gcd(ll a, ll b) {return b ? gcd(b, a%b) : a;}  // Greatest Common Divisor ユークリッドの互除法（aが長辺）
vector<pair<ll,int>> factorize(ll n) {            // a,bの公約数 = GCD(a, b)の約数
    vector<pair<ll,int>> res;
    for (ll i = 2; i*i <= n; i++) { // √nまで探せばよい
        if (n%i) continue;
        res.emplace_back(i, 0);     // 割り切れたら
        while (n%i == 0) {
            n /= i;
            res.back().second++;
        }
    }
    if (n != 1) res.emplace_back(n, 1); // この時点で1でなかったら、残りは素数.
    return res;
}
ll lcm(ll a, ll b) {return a/gcd(a,b)*b;}         // Least Commont Multiple オーバーフローしないように先に割る
ll ceil(ll a, ll b) {return (a+b-1)/b;}

const ll INF = LONG_MAX;
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(int argc, char* argv[])
{
#ifdef LOCAL
    // 以降 cin の入力元が 'xxx.txt' になる
    // std::ifstream in(argv[1]);
    //std::cin.rdbuf(in.rdbuf());
    //std::ifstream in(string(argv[0]) + ".txt");
    std::ifstream in("in.txt");
    std::cin.rdbuf(in.rdbuf());
#endif

    string s_ori;
    cin >> s_ori;
    int q;
    cin >> q;

    bool d = true;

    string s = "";

    int cnt_l = 0;
    int cnt_r = 0;

    rep(i, q) {
        int q1;
        cin >> q1;
        if (q1 == 1) {
            d = !d;
        }
        else {
            int f;
            cin >> f;
            string c;
            cin >> c;
            if (f == 1) {
                if (d) {
                    s = c + s;
                    cnt_l++;
                }
                else {
                    s = s + c;
                    cnt_r++;
                }
            }
            else {
                if (d) {
                    s = s + c;
                    cnt_r++;
                }
                else {
                    s = c + s;
                    cnt_l++;
                }
            }
        }
        //cout << s << endl;
    }

    if (d) {
        cout << s << "  l,r:" << cnt_l << "," << cnt_r << endl;
        cout << s.substr(0, cnt_l) + s_ori + s.substr(cnt_l, cnt_r);
    }
    else {
        reverse(s_ori.begin(), s_ori.end());
        reverse(s.begin(), s.end());

        swap(cnt_l, cnt_r);

        cout << s << "  l,r:" << cnt_l << "," << cnt_r << endl;
        cout << s.substr(0, cnt_l) + s_ori + s.substr(cnt_l, cnt_r);

    }

    return 0;
}

/*
  - 1e9 : 10^9
  - container一般
    - 合計
      accumulate(a.begin(), a.end(), 0LL)  0LLはオーバーフロー用
  - vector
    vector<int> A(N, 0); // 0で初期化(省略可)
    - loop
      for  (int i = 0; i < A.size(); i++) {}
      for  (int elem : A) {}
    - sort
      std::sort(v.begin(), v.end());  // 昇順
      std::sort(v.begin(), v.end(), std::greater<int>()); //降順
      - vector<vector<int>> cnt(h, vector<int>(w))
    - sort(struct)
        struct st_t {
            string name;
            int p;
            bool operator<(const st_t& another) const {
                if (name == another.name)
                {
                    return p > another.p;
                }
                return name < another.name;
            }
        };
        vector<st_t> st(n);
        sort(st.begin(), st.end());
  - pair
    - pairのソートの定義 : firstが優先的に比較。同じだったらsecond。
    - pair<pair<string,int>,int> p[110];
      std::sort(p,p+a);
        こうすると、first.first -> first.second -> secondの順にソートされる
    - sort
      - sort(a.begin(), cb.end(), greater<pair<int, int>>());
  - map
    - for (auto x : map) {}  // x.first:key  x.second:value.
  - priority_queue<int> q;
    - q.top()に最大値が入っている
  - string
    - loop
      for (char& c : s) {}
 */