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: Time Limit Exceeded)

#include<bits/stdc++.h>
#include<cmath>
#pragma GCC optimize("Ofast")
using namespace std;
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
typedef long long ll;
const ll MOD = 1e9+7;
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REP(i,num,n) for(int i=num, i##_len=(n); i<i##_len; ++i)
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<class T> inline void add(T &a, T b){a = ((a+b) % MOD + MOD) % MOD;};
#define print(x) (cout << (x) << endl)
#define pb push_back
#define mp make_pair
#define sz(x) int(x.size())
#define fs first
#define sc second


const int INF = 1e9;
const ll LLINF = 1e16;

template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}
 
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}
 
template<typename T,typename U,typename... V>
typename enable_if<is_same<T, U>::value!=0>::type
fill_v(U &u,const V... v){u=U(v...);}
 
template<typename T,typename U,typename... V>
typename enable_if<is_same<T, U>::value==0>::type
fill_v(U &u,const V... v){
  for(auto &e:u) fill_v<T>(e,v...);
}
 
template<typename T>
T gcd(T a,T b){return b?gcd(b,a%b):a;}

template<typename T>
T ngcd(vector<T> a){
    T res;
    res = a[0];
    for(int i = 1; i < a.size() && res != 1; i++) {
        res = gcd(a[i], res);
    }
    return res;
}
 
template<typename T>
T lcm(T a, T b) {
    return a / gcd(a, b) * b;
}

template<typename T>
T nlcm(vector<T> numbers) {
    T res;
    res = numbers[0];
    for (int i = 1; i < numbers.size(); i++) {
        res = lcm(res, numbers[i]);
    }
    return res;
}

// 割り算の切り上げ処理
template<typename T>
T round_up(T a, T b) {
  return (a + b - 1) / b;
}

// 約数列挙
template<typename T>
vector<T> calc_divisor(T n) {
   vector<T> res;
   for (T i = 1LL; i*i <= n; ++i) {
       if (n % i == 0) {
           res.push_back(i);
           T j = n / i;
           if (j != i) res.push_back(j);
       }
   }
   sort(res.begin(), res.end());
   return res;
}

// 素数判定
template<typename T>
bool isPrime(T x){
    T i;
    if(x < 2)return 0;
    else if(x == 2) return 1;
    if(x%2 == 0) return 0;
    for(i = 3; i*i <= x; i += 2) if(x%i == 0) return 0;
    return 1;
}

// 桁和
template<typename T>
T digsum(T n) {
    T res = 0;
    while(n > 0) {
        res += n%10;
        n /= 10;
    }
    return res;
}

// 指定した文字cが文字列に何文字入ってるか
ll stringcount(string s, char c) {
    return count(s.cbegin(), s.cend(), c);
}


using pint = pair<int, int>;
using pll = pair<ll, ll>;

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};



int main(void)
{
    cout.tie(0);
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    string s;
    cin >> s;
    int q;
    cin >> q;
    rep(i, q){
        int a;
        cin >> a;
        if(a == 1) reverse(all(s));
        else {
            int f;
            char c;
            cin >> f >> c;
            if(f == 1) s = c + s;
            else s = s + c;
        }
    }
    print(s);
}