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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <list>
#include <ctime>
#include <sstream>
#include <queue>
#include <stack>
#include <bitset>
#include <numeric>
#include <cassert>
using namespace std;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef long long ll;
#define FOR(x, b, e) for(int x=(b); x<=(e); ++x)
#define FORD(x, b, e) for(int x=((int)(b))-1; x>=(e); --x)
#define REP(x, n) for(int x=0; x<(n); ++x)
#define ALL(c) c.begin(),c.end()
#define sz(x) ((int)((x).size()))
#define pb push_back
#define st first
#define nd second
#define mp(x,y) make_pair(x,y)
typedef short int sint;

#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
~debug() { cerr << endl; }
eni(!=) cerr << boolalpha << i; ris; }
eni(==) ris << range(begin(i), end(i)); }
sim, class b dor(pair < b, c > d) {
  ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
  *this << "[";
  for (auto it = d.b; it != d.e; ++it) {
	*this << ((it != d.b) ? ", " : "") << *it;
  }
  ris << "]";
}
#else
sim dor(const c&) { ris; }
#endif
};
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

int main() {
	ios_base::sync_with_stdio(false);
	string x;
	cin >> x;
	int przo = 1;
	int q;
	cin >> q;
	vector<char> przod, tyl;
	while (q--) {
		int t;
		cin >> t;
		if (t == 1) {
			przo = 1 - przo;
		} else {
			int f;
			char c;
			cin >> f >> c;
			--f;
			if (f + przo == 1) przod.pb(c);
			else tyl.pb(c);
		}
	}
	vector<char> odp;
	if (przo == 0) {
		// printf("przo: %d, sz: %d, %d\n", przo, sz(przod), sz(tyl));
		swap(przod, tyl);
		reverse(ALL(x));
		// reverse(ALL(przod));
	}
	FORD(i, sz(przod), 0) cout << przod[i];
	cout << x;
	REP(i, sz(tyl)) cout << tyl[i];
	cout << endl;
}