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;

#define rep(i,n) for(int i=0;i<(n);i++)
#define repa(i,a,n) for(int i=(a);i<(n);i++)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define rrepa(i,a,n) for(int i=n-1;i>=(a);i--)
#define all(a) (a).begin(),(a).end()
#define MOD 1000000007
#define cmax(a,b) a=max(a,b)
#define cmin(a,b) a=min(a,b)
#define vc vector

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<int,int> pi;

int main(){

    string s;
    int q;
    cin>>s>>q;

    vi que_t(q),que_f(q);
    vc<char> s1,s2,que_c(q);
    rep(i,q){
        cin>>que_t[i];
        if(que_t[i]==2){
            cin>>que_f[i]>>que_c[i];
        }
    }
    int rev=1;
    rep(i,q){
        if(que_t[i]==1)rev=!rev;
        else{
            if(que_f[i]-1==rev)s1.push_back(que_c[i]);
            else s2.push_back(que_c[i]);
        }
    }

    if(rev){
        rrep(i,s2.size())cout<<s2[i];
        cout<<s;
        rep(i,s1.size())cout<<s1[i];
        cout<<"\n";
    }else{
        reverse(all(s));
        rrep(i,s1.size())cout<<s1[i];
        cout<<s;
        rep(i,s2.size())cout<<s2[i];
        cout<<"\n";
    }

    return 0;
}