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)

#include <bits/stdc++.h>
#include <math.h>
#define REP(i, n) for(int i = 0; i < (n); i++)
#define ALL(v) (v).begin(), (v).end()
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1001001001;
const int mINF = -1001001001;


int main() {
  char S[1000000];
  vector<string> F_1,F_2;
  int Q;
  cin >> S >> Q;
  int S_size = strlen(S);
  int T,F;
  string C;
  int T1_count = 0;
  REP(i,Q) {
    cin >> T;
    if(T==1) {
      T1_count++;
    } else {
      cin >> F >> C;
      if((F+T1_count)%2==1) {
        F_2.push_back(C);
      } else {
        F_1.push_back(C);
      }
    }
  }
  ll F_1_size = F_1.size();
  ll F_2_size = F_2.size();
  if(T1_count%2==1) {
    for(int j = F_1_size-1;j>=0;j--) cout << F_1[j];
    REP(j,S_size) cout << S[j];
    REP(j,F_2_size) cout << F_2[j];
  } else {
    for(int j = F_2_size-1;j>=0;j--) cout << F_2[j];
    for(int j = S_size-1;j>=0;j--) cout << S[j];
    REP(j,F_1_size) cout << F_1[j];
  }
  cout << endl;
  return 0;
}