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;

typedef long long ll;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl

const int maxn = 1e6 + 5;

deque<char> qq;

int main() {
  ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);

  string s; cin>>s;
  int rev = 0;
  for (char c: s) {
    qq.push_back(c);
  }
  int q; cin>>q;

  while (q--) {
    int typ; cin>>typ;
    if (typ == 1) {
      rev++;
    } else {
      int f; char c; cin>>f>>c;
      if (f == 1) {
	if (rev%2==0) {
	  qq.push_front(c);
	} else {
	  qq.push_back(c);
	}
      } else {
	if (rev%2==0) {
	  qq.push_back(c);
	} else {
	  qq.push_front(c);
	}
      }
    }
  }

  string res;
  for (char c: qq) res += c;
  if (rev%2) reverse(res.begin(), res.end());
  out(res);
  return 0;
}
// don't forget to read in the input :^)