/******************************************************************************
* Sample solution for the Necklace Decomposition problem in NWERC'04
* Author: Andreas Björklund
* Solution: Dynamic programming. Let dyn[i] contain the index such that 
*           S[dyn[i],i-1] is the rightmost necklace in the necklace 
*           decomposition for S[1,i-1]. 
*
******************************************************************************/

#include <stdio.h>
#include <string.h>

#define MAXLEN (200)

int n;
char buf[MAXLEN+1];

int compare(char *A,int alen,char *B, int blen)
{
  int j=0;  
  while (j<alen && j<blen && A[j]==B[j]) j++;
  if (j==alen)
  {
    if (j==blen) return 0; else return -1;
  }
  else if (j==blen)
  {
    return 1;
    
  }
  else if (A[j]<B[j])
  {
    return -1;
  }
  else
  {  
    return 1;
  }
}

int isNecklace(char *streng,int len)
{
  int i,j;
  int who=0;
  for (i=1;i<len && !who;i++)
  {
    int a,b;
    j=0;
    a=who;
    b=i;
    while (j<len && streng[a]==streng[b])
    {
      j++;
      a=(a+1) % len;
      b=(b+1) % len; 
    } 
    if (j<len && streng[a]>streng[b]) who=i;
  }
  return who==0;
}

void recp(char* streng,int* who,int where)
{
  int j;
  if (who[where]==-1) return;
  recp(streng,who,who[where]);
  printf("(");
  for (j=who[where];j<where;j++)
  {
    printf("%c",streng[j]);
  }  
  printf(")");
}

void factor(char *streng,int len)
{
  int i,j;
  int dyn[MAXLEN+1];
  dyn[0]=-1;
  for (i=1;i<=len;i++) dyn[i]=0;
  
  for (i=0;i<len;i++)
  {
    for (j=i;j>=0;j--)
    {
      if (isNecklace(&streng[j],i-j+1))
      {
        if (dyn[j]==-1 || compare(&streng[dyn[j]],j-dyn[j]+1,&streng[j],i-j+1)==1)
        {
          dyn[i+1]=j;  
        }
      }
    }    
  }
  recp(streng,dyn,len);
}

int main(void)
{
  scanf("%d",&n);
  while(n--)
  {
    scanf("%s",buf);
    factor(buf,strlen(buf));
    printf("\n");
  }
  return 0;
}