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)

// #pragma GCC target("avx")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
// #define BEGIN_STACK_EXTEND(size) void * stack_extend_memory_ = malloc(size);void * stack_extend_origin_memory_;char * stack_extend_dummy_memory_ = (char*)alloca((1+(int)(((long long)stack_extend_memory_)&127))*16);*stack_extend_dummy_memory_ = 0;asm volatile("mov %%rsp, %%rbx\nmov %%rax, %%rsp":"=b"(stack_extend_origin_memory_):"a"((char*)stack_extend_memory_+(size)-1024));
// #define END_STACK_EXTEND asm volatile("mov %%rax, %%rsp"::"a"(stack_extend_origin_memory_));free(stack_extend_memory_);

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<unordered_map>
#include<complex>
using namespace std;
// const long long mod=998244353;
// const long long inf=mod*mod;
// const long long d2=(mod+1)/2;
long long mod;
const long double EPS=1e-9;
const long double INF=1e+10;
const long double PI=acos(-1.0);
const int C_SIZE = 3100000;
const int UF_SIZE = 3100000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
	int UF[UF_SIZE];
	void init_UF(int n){
		for(int i=0;i<n;i++)UF[i]=-1;
	}
	int FIND(int a){
		if(UF[a]<0)return a;
		return UF[a]=FIND(UF[a]);
	}
	void UNION(int a,int b){
		a=FIND(a);b=FIND(b);if(a==b)return;
		// if(UF[a]>UF[b])swap(a,b);
		UF[a]+=UF[b];UF[b]=a;
	}
}
// ここから編集しろ
char in[110000];
vector<char>L;
char ch[2];
vector<char>R;
int main(){
	int a;scanf("%s%d",in,&a);
	int fi=0;
	for(int i=0;i<a;i++){
		int t;scanf("%d",&t);
		if(t==1){
			fi=!fi;
		}else{
			int f;scanf("%d",&f);
			scanf("%s",ch);
			if((fi+f)%2==0){
				R.push_back(ch[0]);
			}else{
				L.push_back(ch[0]);
			}
		}
	}
	if(fi){
		for(int i=0;i<R.size();i++)printf("%c",R[R.size()-1-i]);
			int n=strlen(in);
			reverse(in,in+n);
		printf("%s",in);
		for(int i=0;i<L.size();i++)printf("%c",L[i]);
		printf("\n");
	}else{
		for(int i=0;i<L.size();i++)printf("%c",L[L.size()-1-i]);
		printf("%s",in);
		for(int i=0;i<R.size();i++)printf("%c",R[i]);
		printf("\n");
	}

}