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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)


import java.io.*;
import java.text.*;
import java.util.*;

public class Main {

    static InputReader sc = new InputReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) throws IOException {
        solver s = new solver();
        int t = 1;
        while (t > 0) {
            s.solve();
            t--;
        }
        out.close();
    }

    static class InputReader {

        public BufferedReader br;
        public StringTokenizer token;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            token = null;
        }

        public String sn() {
            while (token == null || !token.hasMoreTokens()) {
                try {
                    token = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return token.nextToken();
        }

        public int ni() {
            return Integer.parseInt(sn());
        }

        public String snl() throws IOException {
            return br.readLine();
        }

        public long nlo() {
            return Long.parseLong(sn());
        }

        public double nd() {
            return Double.parseDouble(sn());
        }

        public int[] na(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.ni();
            return a;
        }

        public long[] nal(int n) {
            long a[] = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.nlo();
            return a;
        }
    }

    static int inf = 5000013;

    static class solver {
        DecimalFormat df = new DecimalFormat("0.00000000");
        long max = 1000000000000000l, min = -10000000000000000l;
        long mod = (long) 1e9 + 7;

        long pro(long a, long b) {
            return (a % mod * b % mod) % mod;
        }

        long add(long a, long b) {
            return (a + b) % mod;
        }

        long sub(long a, long b) {
            return ((a % mod - b % mod) + mod) % mod;
        }

        long ncr3(long n) {
            return (n * (n - 1) * (n - 2)) / 6l;
        }

        long ncr2(long n) {
            return n * (n - 1) / 2l;
        }

        int modPow(int x, int y) {
            int res = 1;
            while (y > 0) {
                if (y % 2 != 0)
                    res *= x;
                x = x * x;
                y = y >> 1;
            }
            return res;
        }

        int gcd(int a, int b) {
            if (b == 0)
                return a;
            return gcd(b, a % b);
        }

        final int maxn = 1000005;

        void solve() {
            char s[] = sc.sn().toCharArray();
            int n = s.length;
            Deque<Character> dq = new LinkedList<>();
            Stack<Character> stack = new Stack<>();
            for (int i = 0; i < n; i++)
                dq.addLast(s[i]);
            int q = sc.ni();
            int flag = 0;
            while (q-- > 0) {
                int t = sc.ni();
                if (t == 1) {
                    flag = (flag + 1) % 2;
                } else {
                    if (flag == 1) {
                        while (!dq.isEmpty())
                            stack.push(dq.pollFirst());
                        while (!stack.isEmpty())
                            dq.addLast(stack.pop());
                        flag = 0;
                    }
                    int f = sc.ni();
                    char c = sc.sn().charAt(0);
                    if (f == 1)
                        dq.addFirst(c);
                    else
                        dq.addLast(c);
                }
            }
            if (flag == 1) {
                while (!dq.isEmpty())
                    stack.push(dq.pollFirst());
                while (!stack.isEmpty())
                    dq.addLast(stack.pop());
            }
            while (!dq.isEmpty()) {
                out.print(dq.pollFirst());
            }
        }
    }
}