Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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 C# (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication14//abcdefghij
{
    class Program
    {
        static void Main()
        {
            string a = Console.ReadLine();
            char[] sss = new char[a.Length];
            for (int r = 0; r < a.Length; r++)
            {
                sss[r] = a[r];
            }
            int b = int.Parse(Console.ReadLine());
            for (int c = 0; c < b; c++)
            {
                string[] s = Console.ReadLine().Split(); int f = int.Parse(s[1]), g = int.Parse(s[2]);
                switch (s[0][2])
                {
                    case 'i':
                        for (int i = f; i < g + 1; i++)
                        {
                            Console.Write(sss[i]);
                        }
                        Console.WriteLine();
                        break;
                    case 'v':
                        char[] sssss = new char[g - f + 1];//5
                        for (int z = f; z < g + 1; z++)
                        {
                            sssss[z - f] = sss[z];
                        }
                        for (int d = f; d < g + 1; d++)
                        {
                            sss[d] = sssss[d+f];
                        }
                        break;
                    case 'p':
                        for (int h = f; h < g + 1; h++)
                        {
                            sss[h] = s[3][h - f];
                        }
                        break;
                }

            }
        }
    }
}