技術標籤：ACM

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

abcd
aaaa
ababab
.

Output

1
4
3

Hint

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

#include <stdio.h>
#include <string.h>

char s[1000001];
int next[1000001];
void getNext(int len)
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])next[++i]=++j;
        else j=next[j];
    }
}
int main()
{
    while(scanf("%s",s)&&s[0]!='.')
    {
        int len=strlen(s);
        getNext(len);
        if(len%(len-next[len])==0)
            printf("%d\n",len/(len-next[len]));
        else printf("1\n");
    }

    return 0;
}