Power Strings

Time Limit: 3000MS		Memory Limit: 65536K

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcdaaaaababab.

Sample Output

143

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

求最短循环节的出现次数

求最短循环节长度的方法:

a=len-next[len]

若len%a==0 ，最短循环节长度为a

否则，没有循环节

#include
      
       #include
       
        #include
        
         using namespace std;int n;char s[1000001];int f[1000010];void getnext(){    for(int i=1;i
         
          >s)    {        if(s[0]=='.') break;        n=strlen(s);        getnext();        int ans=n-f[n];        if(n%ans) printf("1\n");        else printf("%d\n",n/ans);    }    }