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: Runtime Error)

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <utility>
#include <set>
#include <stack>
#include <climits>
#include <cmath>

#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define rep1(i,n) for(int i=1, i##_len=(n); i<=i##_len; ++i)
#define reps(i,s,n) for(int i=s, i##_len=(n); i<i##_len; ++i)
#define all(x) (x).begin(),(x).end()
#define MEMSET(v, h) memset((v), h, sizeof(v))
//#include <boost/multiprecision/cpp_dec_float.hpp>
//#include <boost/multiprecision/cpp_int.hpp>
typedef long long ll;
const char sp = ' ';


using namespace std;

int main(void){
    ll N, p;
    ll dp10[300000], dpS[300000];
    ll kosu[10001];
    MEMSET(kosu, 0);
    string S;
    cin>>N>>p;
    cin>>S;
    dp10[0]=1; dpS[0]=(S[N-1]-'0')%p;
    if(p==2){
        ll count=0;
        rep(i, N){
            ll th=S[N-1-i];
            if((S[i]-'0')%2==0) count+=(N-i);
        }
        cout<<count<<endl;
        return 0;
    }
    if(p==5){
        ll count=0;
        rep(i, N){
            ll th=S[N-1-i];
            if((S[i]-'0')%5==0) count+=(N-i);
        }
        cout<<count<<endl;
        return 0;
    }
    //cout<<dpS[0]<<endl;
    rep(i, N){
        ll th=S[N-i-1]-'0';
        //cout<<th<<endl;
        if(i==0) continue;
        dp10[i]=(dp10[i-1]*10)%p;
        dpS[i]=(dpS[i-1]+th*dp10[i])%p;
        //cout<<dpS[i]<<sp<<dp10[i]<<endl;
    }
    rep(i, N){
        kosu[dpS[i]]++;
    }
    //cpp_int count=0;
    ll count=0;
    rep(i, p){
        count+=(kosu[i]*(kosu[i]-1))/2;
        //cout<<i<<":"<<sp<<kosu[i]<<endl;
    }
    count+=kosu[0];
    cout<<count<<endl;
    return 0;
}