/* Sample solution to Necklace
 * Author: Mikael Goldmann
 *
 * Checks input (does not check for garbage after last case)
 *    No leading or trailing blanks allowed for any case.
 *
 * Problem solved by necklace() and isNecklace(). 
 * Observe (or guess) that we should always choose the longest prefix
 * that is a necklace to be the first in the decomposition. 
 * Decompose remaning suffix recursively.
 */

import java.io.*;

class necklace_mg {
    static BufferedReader stdin;
    static PrintStream stdout;

    public static void main(String[] a) 
    {
	Reader rdr = new InputStreamReader(System.in);
	stdin = new BufferedReader(rdr);
	OutputStream os = new BufferedOutputStream(System.out);
	stdout =  new PrintStream(os);

	(new necklace_mg()).run();

	stdout.close();
    }

    void run() 
    {
	int i = 0, n = 0;
	String s = "";
	try {
	    i = 0;
	    s = stdin.readLine();
	    n = Integer.parseInt(s);	
	    for (i = 1; i <= n; ++i)
		solve(stdin.readLine(), i);
	}
	catch (NumberFormatException e) {
	    System.err.println("Number of cases (" + s 
			       +  ") is not an integer " 
			       + "(or contains blanks)"); 
	}
	catch (IOException e) {
	    if (i == 0)
		System.err.println("Cannot read number of cases");
	    else 
		System.err.println("Cannot read case " + i);
	}
    }

    void solve(String s, int num) 
    {
	/* start input validity check */
	if (s == null) {
	    System.err.println("Case " + num + ": null string");
	    return;
	}
	int n = s.length();
	if (n == 0) {
	    System.err.println("Case " + num + ": empty string");
	    return;
	}
	for (int i = 0; i < n; ++i)
	    if (s.charAt(i) != '0' && s.charAt(i) != '1') {
		System.out.println("Case " + num 
				   + ": illegal character ("
				   + s.charAt(i) + ") at index"
				   + i);
		return;
	    }
	/* end check */

	necklace(s);
    }
    
    void necklace(String s)
    {
	int n = s.length();
	if (n == 0)
	    stdout.println();
	else {
	    while (!isNecklace(s.substring(0, n)))
		--n;
	    stdout.print("(" + s.substring(0,n) + ")");
	    necklace(s.substring(n));
	}
    }
    
    boolean isNecklace(String s)
    {
	String p = s+s;
	int n = s.length();
	for (int k = 1; k < n; ++k){
	    for (int i = 0; i < n; ++i) {
		if (p.charAt(i) < p.charAt(i+k)) break;
		else if (p.charAt(i) > p.charAt(i+k)) return false;
	    }
	}
	return true;
    }
}