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;
typedef pair<ll, ll> P;
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define IFOR(i, m, n) for (ll i = n - 1; i >= m; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define FOREACH(x, a) for (auto&(x) : (a))
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) ll(x.size())

int main() {
  string s;
  cin >> s;
  ll q;
  cin >> q;
  deque<ll> dq;
  REP(i, SZ(s)) { dq.push_back(s[i] - 'a'); }
  bool reversed = false;
  while (q--) {
    ll type;
    cin >> type;
    if (type == 1) {
      reversed = !reversed;
      continue;
    }
    ll f;
    char c;
    cin >> f >> c;
    f--;
    if (!(reversed ^ f)) {
      dq.push_front(c - 'a');
    } else {
      dq.push_back(c - 'a');
    }
  }

  if (reversed) {
    IREP(i, SZ(dq)) { cout << (char)(dq[i] + 'a'); }
  } else {
    REP(i, SZ(dq)) { cout << (char)(dq[i] + 'a'); }
  }
  cout << endl;
}