#include<stdio.h>
#include <string.h>
#define maxn 1000010
char a[maxn];
int nxt[maxn];
int len;
void getnxt()
{
	int i=1,j=0;
	nxt[0]=0;
	 while(i<len)
	 {
	 	if(a[i]==a[j])
	 	{
	 		nxt[i++]=++j;
		 }
		 else if(!j)
		 {
		 	i++;
		 }
		 else 
		 {
		 	j=nxt[j-1];	
		 }
	 }
}
int main()
{
	while(scanf("%s",a)!=EOF)
	{
	 memset(nxt,0,sizeof(nxt));//我錯誤的原因是沒有給nxt清零 
		if(strcmp(a,".")==0)
		{
			break;
		}
		len=strlen(a);
		getnxt();
		int sum=len-nxt[len-1];
	 	if(len%sum!=0) printf("1\n");
		else printf("%d\n",len/sum) ;
 
	}
	return 0;
}
 

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

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

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