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: Runtime Error)

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

int main() {
	char S[300001], *SP, C, tmp;
	unsigned int Q, i, j, x;
	unsigned short T, F;
	char isReverse = 0;

	memset(S, 0, sizeof(S));
	SP = &S[100000];
	scanf("%s", SP);
	scanf("%u", &Q);
	for (i = 0; i < Q; i++) {
		scanf("%hu", &T);
		if (T == 1) {
			if (isReverse == 2) {
				isReverse -= 2;
			} else {
				isReverse += 2;
			}
		} else {
			scanf("%hu %c", &F, &C);
			if (F + isReverse == 1
					|| F + isReverse == 4) {
				*(--SP) = C;
			} else {
				SP[strlen(SP)] = C;
			}
		}
	}
	if (isReverse == 2) {
		for (j = strlen(SP) - 1; strlen(SP) / 2 <= j; j--) {
			tmp  = SP[j];
			SP[j] = SP[x];
			SP[x] = tmp;
			x++;
		}
	}
	printf("%s\n", SP);
	return 0;
}