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)

/*by freesteed*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define fill(x,c) memset(x,c,sizeof(x))
#define fi first
#define se second
#define sz(A) (int)(A.size())
#define all(x) (x).begin(),(x).end()
#define caset int ___T; cin>>___T; for(int cs=1;cs<=___T;cs++)
#define get_be(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
//get_be biger & equal
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
const ll mod=1e9+7;
const ll oo=1e9+10;
const int Max = 1e6+10;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll a[Max],b[Max];
ll n,m,k,res; 
struct BIT {
private:
	vector<ll> node; int n;
public:
	void init(int n_) {
		n = n_; node.resize(n, 0);
	}
	//0-indexed
	void add(int a, ll w) {
		for (int i = a; i < n; i |= i + 1)node[i] += w;
	}
	//[0,a)
	ll sum(int a) {
		ll ret = 0;
		for (int i = a - 1; i >= 0; i = (i&(i + 1)) - 1)ret += node[i];
		return ret;
	}
	//[a,b)
	ll sum(int a, int b) {
		return sum(b) - sum(a);
	}
};
void solve(){
	string s;
	cin>>s;
	ll q;cin>>q;
	ll flag = 1,cnt=0;
	string front="",back="";
	while(q--){
		ll tep;
		cin>>tep;
		if(tep==1){
			cnt++;
			if(flag == 1)flag = 2;
			else flag = 1;
		}else{
			ll p;
			char c;
			cin>>p>>c;
			if(p==flag)front+=c;
			else back+=c;
		}
	}
	
	if(cnt&1){
		cout<<front;
		for(int i=sz(s)-1;i>=0;i--)cout<<s[i];
		for(int i=sz(back)-1;i>=0;i--)cout<<back[i];
	}
	else {
		for(int i=sz(front)-1;i>=0;i--)cout<<front[i];cout<<s;
		cout<<back; 
	}
	
}
int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	//caset{
		solve();
	//}
  	return 0;
}