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 <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vii = vector<vi>;
using in6 = int64_t;
using ind = double;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int main() {
  string s;
  cin >> s;
  int q;
  cin >> q;
  int size = s.size();
  vector<char> data(50 * q);
  int left = 25 * q;
  int right = left + size - 1;
  rep(i,size) {
    data[left + i] = s[i];
  }
  rep(i,q) {
    int a;
    cin >> a;
    if(a == 1) {
      int n,m;
      n = right;
      m = left;
      right = m;
      left = n;
    }
    else {
      size++;
      int b;
      char c;
      cin >> b >> c;
      if(b == 1) {
        if(left <= right) {
          data[left - 1] = c;
          left--;
        }
        else {
          data[left + 1] = c;
          left++;
        }
      }
      else {
        if(left <= right) {
          data[right + 1] = c;
          right++;
        }
        else {
          data[right - 1] = c;
          right--;
        }
      }
    }
  }
  if(left <= right) {
    rep(i,size) {
      cout << data[left + i];
    }
  }
  else {
    rep(i,size) {
      cout << data[left - i];
    }
  }
  cout << endl;
}