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 vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define REP(i, a, b) for(ll i = a; i < b; i++)
#define out(x) std::cout << x << '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define pb push_back
#define sz(x) ((int)(x).size())
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { ll g = gcd(a, b); return a / g * b; }
const int dx[4] = { 1, 0, -1,  0 };
const int dy[4] = { 0, 1,  0, -1 };
const int INF = INT_MAX / 2;
const ll  inf = LLONG_MAX / 2;
const int mod = 1000000007;
const int MAX_N = 101010;
const long double PI = acos(-1);

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(10);
	/*------------------------------*/

	string s;
	cin >> s;

	deque<char> ds(sz(s));
	rep(i, sz(s)) {
		ds[i] = s[i];
	}

	int q;
	cin >> q;

	int rev = 0;
	rep(i, q) {
		int type;
		cin >> type;
		if (type == 2) {
			int f;
			char c;
			cin >> f >> c;
			f--;
			if (rev) {
				f = 1 - f;
			}
			if (f == 0) {
				ds.push_front(c);
			}
			else {
				ds.push_back(c);
			}
		}
		else {
			rev = 1 - rev;
		}
	}

	if (rev) {
		reverse(ds.begin(), ds.end());
	}

	for (auto c : ds) {
		cout << c;
	}
	cout << endl;

	return 0;
}