Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 50627		Accepted: 21118

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.

Source

 

求最短循环节出现次数

1 #include 
      
        2 #include 
       
         3 #include 
        
          4  5 using namespace std; 6  7 const int N(2333333); 8 int ans,len,p[N]; 9 char s[N];10 11 inline void Get_next()12 {13     for(int i=2,j=0;i<=len;i++)14     {15         for(;s[i]!=s[j+1]&&j>0;) j=p[j];16         if(s[i]==s[j+1]) j++;17         p[i]=j;18     }19 }20 21 int main()22 {23     for(scanf("%s",s+1);s[1]!='.';scanf("%s",s+1))24     {25         memset(p,0,sizeof(p));26         len=strlen(s+1);27         Get_next();28         int tmp=len-p[len];29         if(len%tmp) puts("1");30         else printf("%d\n",len/tmp);31     }32     return 0;33 }