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)

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define ff first
#define ss second
#define file_se_input  freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#define pp long long int
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define pb push_back
#define mp make_pair
#define pi 3.141592653589793238
#define eb emplace_back
#define rep(i,a,b) for (int i = a; i <= b; i++)
#define zip(i,a,b) for(int i=a;i<b;i++)
#define rzip(i,a,b) for(int i=a;i>=b;i--)
#define ll unsigned long long int
#define test int t;cin>>t; while(t--)
#define um unordered_map
#define en '\n'
#define us unordered_set
typedef pair<int, int>     pii;
typedef pair<char, int>    pci;
typedef pair<char, char>   pcc;
typedef vector<pii>       vii;
typedef long double ld;
#define all(v) v.begin(), v.end()
#define INF (1e18+5)
#define inf (1e9+5)
#define mod 1000000007


bool check_prime(long long n)
{
	int flag = 0;
	for (long long i = 2; i * i <= n; i++)
	{
		if (n % i == 0)
		{
			flag = 1;
			break;
		}
	}
	if (n == 1)
		return false;
	else if (flag == 0 || n == 2 || n == 3)
	{
		return true;
	}
	else
	{
		return false;
	}

}

int BE(int x, int n, int m) //function to calculate x raise to power n modulo m
{
	int result = 1;
	while (n > 0)
	{
		if (n % 2 == 1)
			result = result * x % m;
		x = x * x % m;
		n = n / 2;
	}
	return result;
}

int32_t main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(0);
#ifndef ONLINE_JUDGE
	if (fopen("input.txt", "r"))
	{
		freopen ("input.txt" , "r" , stdin);
		freopen ("output.txt" , "w" , stdout);
	}
#endif
	//_______________________________-code starts-_______________________________________________

	string s;
	cin >> s;
	string a = "12";
	int q;
	cin >> q;
	int t;
	int f;
	char c;
	string x;
	int var = 0;
	zip(i, 0, q)
	{
		cin >> t;
		if (t == 1)
		{
			//reverse(all(a));
			var++;
		}
		else
		{
			cin >> f >> c;
			if (f == 1)
			{
				x = c;
				if (var % 2)
				{
					a = x + a;
				}
				else
				{
					a = a + x;
				}
			}
			else
			{
				x = c;
				if (var % 2)
				{
					a = a + x;
				}
				else
				{
					a = x + a;
				}
			}
		}
	}
	if (var % 2 == 0)
	{
		reverse(all(a));
	}
	string y = s;
	reverse(all(y));
	for (int i = 0; i < a.size(); i++)
	{
		if (a[i] == '1' || a[i] == '2')
		{
			if (a[i] == '1')
			{
				cout << s;
				i++;
			}
			else
			{
				cout << y;
				i++;
			}
		}
		else
		{
			cout << a[i];
		}
	}
	cout << en;


	return 0;
}