#include<iostream>

 #include<cstring>

 #include<cstdio>

 using namespace std;

 const int N = , INF = 1e9;

 struct node{

     char s[N];  //dp[i][j]的len表示区间i-j压缩后的最小长度 s表示压缩后的字串

     int len;

 }dp[N][N];

 int n;

 char s[N];

 int main(){

     scanf("%s", s + );

     n = strlen(s + );

     for(int i = ; i <= n; i++)

       dp[i][i].len = , dp[i][i].s[] = s[i];

     for(int l = ; l <= n; l++)

       for(int i = ; i + l -  <= n; i++){ //枚举区间

           int j = i + l - ;

           dp[i][j].len = INF;

           for(int nowl = ; nowl <= l / ; nowl++){  //枚举一个区间所有可能被压缩后子串的长度

               if(l % nowl != ) continue;  //如果无法整除肯定不能被压缩

               int st = i, ed = i + nowl;

               while(s[st] == s[ed] && ed <= j) st++, ed++;  //检验该串能否被压缩

               if(ed > j){

                   int num = l / nowl;  //压缩后串的数目为num

                   sprintf(dp[i][j].s, "%d", num);  //那么该状态下最小串的开头就是num

                   strcat(dp[i][j].s, "(");

                   strcat(dp[i][j].s, dp[i][i + nowl - ].s); //把压缩后的串接起来

                   strcat(dp[i][j].s, ")");

                   dp[i][j].len = strlen(dp[i][j].s);

                   break;

               }

           }

           for(int k = i; k < j; k++)

             if(dp[i][j].len > dp[i][k].len + dp[k + ][j].len){  //重新扫一遍区间更新dp[i][j]

                 dp[i][j].len = dp[i][k].len + dp[k + ][j].len;

                 strcpy(dp[i][j].s, dp[i][k].s);

                 strcat(dp[i][j].s, dp[k + ][j].s);

             }

       }

     puts(dp[][n].s);

     return ;

 }