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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int acs(const void *a, const void *b) { 
  return *(int*)a - *(int*)b;
} /* 1,2,3,4.. */
int des(const void *a, const void *b) {
  return *(int*)b - *(int*)a;
} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b) { 
  return *(char*)a - *(char*)b;
} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b) {
  return strcmp(*(const char **)a, *(const char **)b);
} /* aaa,aab.. */
#define min(a,b) (a < b ? a : b)
#define max(a,b) (a > b ? a : b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 100005
#define MOD 1000000007
#define INF 1000000009
typedef long long int lli;

int main(void) {
  char s[MAX];
  scanf("%s", s);
  int q;
  scanf("%d", &q);
  int t, f;
  char c;

  int si = strlen(s);

  char head[2*MAX] , rear[2*MAX];
  int hi = 2*MAX-1, ri = 0;
  int rev = 0;

  rep(i, 0, q) {
    scanf("%d", &t);
    if ( t == 1 ) rev = abs(rev-1);
    else {
      scanf("%d %c", &f, &c);
      if ( f == 1 ) {
        if ( rev ) { // 反転している
          rear[ri++] = c;
        } else {
          head[hi--] = c;
        }
      } else {
        if ( rev ) { // 反転している
          head[hi--] = c;
        } else {
          rear[ri++] = c;
        }
      }
    }
  }
  if ( rev ) {
    for ( int i = ri-1; i >= 0; i-- ) 
      printf("%c", rear[i]);
    for ( int i = si-1; i >= 0; i-- ) 
      printf("%c", s[i]);
    for ( int i = 2*MAX-1; i > hi; i-- ) 
      printf("%c", head[i]);
  } else {
    for ( int i = hi+1; i <= 2*MAX; i++ ) 
      printf("%c", head[i]);
    for ( int i = 0; i < si; i++ ) 
      printf("%c", s[i]);
    for ( int i = 0; i < ri; i++ ) 
      printf("%c", rear[i]);
  }
  printf("\n");
  return 0;
}