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)

/**
 * author:    Ziel
 * created:   11.04.2020 22:17:38
**/

// ~zaletaet~

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>

#define fi first
#define se second
#define sz(x) int(x.size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin, x.rend()
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define count1 __builtin_popcountl
#define debug(x) cerr << (#x) << " = " << (x) << "\n"
#define ACCELERATE ios_base::sync_with_stdio(false),cin.tie(nullptr)
#define fre(f) if(fopen(f".in", "r")) freopen(f".in", "r", stdin),freopen(f".out", "w", stdout)

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

using ll = long long;
using ull = unsigned long long;
// using big = __int128_t; // -10^38...10^38
using db = double;
using ld = long double;
typedef pair<int,int> ii;
typedef pair<ll,ll> pll;

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class T>
using p_queue = priority_queue<T, vector<T>, greater<T>>;
template <class T>
inline void cmin(T &a, T b) { a = min(a, b); }
template <class T>
inline void cmax(T &a, T b) { a = max(a, b); }

const ll N = 1e5 + 123, MOD = 1e9 + 7, oo = 1e9 + 9, INF = 1e18 + 9;
const db EPS = 1e-9, PI = acos(-1); // 3.14159265358979323846
const int dx[] = {1, -1, 0, 0, 1, 1, -1, -1}, dy[] = {0, 0, 1, -1, 1, -1, -1, 1};

main() {
#ifdef ZVER
	cerr << "Ready\n";
#endif
	fre("");
	ACCELERATE;
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	
	string s;
	cin >> s;
	
	int q;
	cin >> q;
	
	bool rev = 0;
	deque<char> bgn, end;
	while (q--) {
		int type;
		cin >> type;
		
		if (type == 1) {
			rev ^= 1;
			reverse(all(bgn));
			reverse(all(end));
			swap(bgn, end);
		}
		else {
			int f;
			char c;
			cin >> f >> c;
			
			if (f == 1) bgn.push_front(c);
			else end.pb(c);
		}
	}
	
	if (rev) reverse(all(s));
	for (char i: bgn) cout << i;
	cout << s;
	for (char i: end) cout << i;
	
	return 0;
}

// uniform_int_distribution<int>(l, r)(rng) -> random in range [l, r]
// assert(condition) -> if (condition == false) Runtime Error
// order_of_key(x) -> number of elements smaller than x
// *find_by_order(k) -> k-th element in 0-based indexation