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 .

Sample Output

1 4 3

 找一个串的最小循环节的个数.。

先求出next数组，然后使用长度-next【len】就是最小循环节的长度。然后特判就行。

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#include<math.h>
using namespace std;
typedef long long ll;
typedef double ld;
char p[1000010 ];
int m,nxt[1000010 ];

void GetNext(char* p, int nxt[])
{
   int pLen=strlen(p);
   nxt[0]=-1;
   int k=-1;
   int j=0;
   while(j<pLen)
   {
      //p[k]表示前缀，p[j]表示后缀
      if(k==-1||p[j]==p[k])
      {
         ++j;
         ++k;
         //较之前next数组求法，改动在下面4行
         if(p[j]!=p[k])
            nxt[j]=k; //之前只有这一行
         else
            //因为不能出现p[j] = p[ next[j ]]，所以当出现时需要继续递归，k = next[k] = next[next[k]]
            nxt[j]=nxt[k];
      }
      else
      {
         k=nxt[k];
      }
   }
}

int main()
{
   while(scanf("%s",p)!=EOF)
    {
        if(p[0]=='.')
          break;
        int len=strlen(p);
        GetNext(p,nxt);
        int ans=1;
        if(!(len%(len-nxt[len])))
        ans=len/(len-nxt[len]);
        printf("%d\n",ans);
    }
    return 0;
}