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.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;

public class Main {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		String S = sc.next();

		StringBuilder sb = new StringBuilder();
		sb.append(S);

		int Q = sc.nextInt();

		for (int i = 0; i < Q; i++) {
			int T = sc.nextInt();
			if (T == 1) {
				sb.reverse();
			} else if (T == 2) {
				int F = sc.nextInt();
				if (F == 1) {
					String C = sc.next();
					sb.insert(0, C);
				} else if (F == 2) {
					String C = sc.next();
					sb.append(C);
				}
			}
		}
		System.out.println(sb);
	}

	//以下、自作ライブラリ

	/*
		int N = sc.nextInt();

		int[] a = new int[N];
		for (int i = 0; i < N; i++)
			a[i] = sc.nextInt();

		String s = sc.next();

		String[] a = new String[N];
		for (int i = 0; i < N; i++)
			a[i] = sc.next();

		int[][] a = new int[N][2];
		for (int i = 0; i < N; i++) {
			a[i][0] = sc.nextInt();
			a[i][1] = sc.nextInt();
		}
		Arrays.sort(a, (x, y) -> Integer.compare(x[0], y[0]));

		List<Integer> list = new ArrayList<>();

		Set<Integer> set = new HashSet<>();

		Map<Integer, Integer> map = new HashMap<>();

		nextLineのときは
		Scanner sc = new Scanner(System.in);
	 */

	//繰り返し二乗法による,べき乗計算で剰余を返す
	public static long modPow(long x, long n, final long mod) {
		long sum = 1;
		while (n > 0) {
			if ((n & 1) == 1) { //ビットが立っていたら
				sum = (sum * x) % mod; //x^2^k1 * x^2^k2 * x^2^k3 * ・・・
			}
			x = (x * x) % mod; //((x^2)^2)^2 ・・・
			n >>= 1;
		}
		return sum;
	}

	//組み合わせの数nCrを計算し剰余を返す
	public static long calcNumOfCombinationMod(long n, int r, final long mod) {
		long num = 1;
		for (int i = 0; i < r; i++) {
			num = (num * (n - i)) % mod;
			num = (num * modPow(i + 1, mod - 2, mod)) % mod;
		}
		return num;

	}

	//組み合わせの数nCrを計算
	public static int calcNumOfCombination(int n, int r) {
		int num = 1;
		for (int i = 1; i <= r; i++) {
			num = num * (n - i + 1) / i;
		}
		return num;
	}

	//階乗
	public static int fact(int n) {
		if (n == 1 || n == 0)
			return 1;
		else
			return n * fact(n - 1);
	}

	//指定した値以上の要素が最初に現れる位置
	//要素の数は+1（配列のズレ)
	public static long lowerBound(int n, long[] x, long y) {
		int left = 0;
		int right = n;
		while (left < right) {
			int mid = (left + right) / 2;
			if (x[mid] < y) {
				left = mid + 1;
			} else {
				right = mid;
			}
		}
		return left;
	}

	//指定した値より大きい要素が最初に現れる位置
	//要素の数は+1（配列のズレ)
	public static long upperBound(int n, long[] x, long y) {
		int left = 0;
		int right = n;
		while (left < right) {
			int mid = (left + right) / 2;
			if (x[mid] <= y) {
				left = mid + 1;
			} else {
				right = mid;
			}
		}
		return left;
	}

	//最大公約数
	public static long gcd(long a, long b) {
		if (a < b)
			return gcd(b, a);
		long c = 0;
		c = a % b;
		if (c == 0)
			return b;
		return gcd(b, c);
	}

	//最小公倍数
	public static long lcm(long m, long n) {
		return m * n / gcd(m, n);
	}

	//素数判定
	public static boolean isPrime(int n) {
		if (n < 2)
			return false;
		if (n == 2)
			return true;
		if (n % 2 == 0)
			return false;

		for (int i = 3; i * i <= n; i += 2) {
			if (n % i == 0)
				return false;
		}
		return true;
	}
}

//以下、高速スキャナークラス
class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	public boolean hasNext() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
		return hasNextByte();
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
			throw new NumberFormatException();
		return (int) nl;
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}
}