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 (GCC 5.4.1) (judge verdict: Accepted)

// ABC 158-D
// 2020.3.7 bal4u

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define gc() getchar_unlocked()
#define pc(c) putchar_unlocked(c)

int in() {   // 非負整数の入力
	int n = 0; int c;
	c = gc();
	do n = 10 * n + (c & 0xf); while ((c = gc()) >= '0');
	return n;
}

int ins(char *s) { // 文字列の入力　スペース以下の文字で入力終了
	char *p = s--;
	do *++s = gc();
	while (*s > ' ');
	*s = 0;
	return s - p;
}

char S[600005], *a, *b;

int main()
{
	int Q, rev, id, F, C;

	a = S+250000;
	b = a + ins(a);
	rev = 0;
	Q = in();
	while (Q--) {
		id = gc(), gc();
		if (id == '1') rev = !rev;
		else {
			F = gc(), gc();
			C = gc(), gc();
			if (F == '1') {
				if (rev) *b++ = C;
				else *--a = C;
			} else {
				if (rev) *--a = C;
				else *b++ = C;
			}
		}
	}
	if (rev) {
		while (b > a) pc(*--b);
	} else {
		while (a < b) pc(*a++);
	}
	pc('\n');
	return 0;
}