Power Strings

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.

让求字符串的最小循环节有几个

#include<stdio.h>#include<math.h>#include<string.h>#include<iostream>#include<algorithm>const int maxn=1e6+10;char s[maxn];int Next[maxn];void getNext(){    int i=0;    int j=-1;    int l=strlen(s);    while(i<l)    {        if(j==-1||s[i]==s[j])        {            i++;            j++;            Next[i]=j;        }        else j=Next[j];    }}int main(){    while(~scanf("%s",s))    {       int l=strlen(s);       if(s[0]=='.')        break;        Next[0]=-1;        getNext();        int ans=l-Next[l];//最小循环节长度。        if(l%ans!=0)//串不是由最小循环节构成            printf("1\n");        else            printf("%d\n",l/ans);    }}