Problem:
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Transformation

Write a program which performs a sequence of commands to a given string $str$. The command is one of:

print a b: print from the a-th character to the b-th character of $str$

reverse a b: reverse from the a-th character to the b-th character of $str$

replace a b p: replace from the a-th character to the b-th character of $str$ with p

Note that the indices of $str$ start with 0.

Input

In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.

Output

For each print command, print a string in a line.

Constraints

$1 \leq $ length of $str \leq 1000$

$1 \leq q \leq 100$

$0 \leq a \leq b < $ length of $str$

for replace command, $b - a + 1 = $ length of $p$

Sample Input 1

abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4

Sample Output 1

xaze

Sample Input 2

xyz
3
print 0 2
replace 0 2 abc
print 0 2

Sample Output 2

xyz
abc

Solution in JAVA (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] arges) {

		Scanner sc = new Scanner(System.in);
		String line = sc.nextLine();
		int v = Integer.parseInt(sc.nextLine());

		List<String> order = new ArrayList<String>();

		for (int i = 0; i < v; i++) {
			order.add(sc.nextLine());
		}

		for (String string : order) {
			String[] orders = string.split(" ");
			int v1 = Integer.parseInt(orders[1]);
			int v2 = Integer.parseInt(orders[2]);
			if (orders[0].equals("replace")) {
				StringBuffer sb = new StringBuffer(line.substring(v1, v2+1));
				sb.replace(0,v2-v1+1, orders[3]);
				line=line.replace(line.substring(v1, v2+1), sb);
			}
			if (orders[0].equals("reverse")) {
				StringBuffer sb = new StringBuffer(line.substring(v1, v2+1));
				sb.reverse();
				line=line.replace(line.substring(v1, v2+1), sb);
			}
			if (orders[0].equals("print")) {
				System.out.println(line.substring(v1, v2+1));
			}
		}
	}
}