Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 39658		Accepted: 16530

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

KMP算法

设字符串的长度为l，假设l%(l-next[l])=0。该序列为循环序列，循环节长度为l-next[l]，答案即为l/(l-next[l])；反之则不为循环序列。答案为1。

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           #define F(i,j,n) for(int i=j;i<=n;i++)#define D(i,j,n) for(int i=j;i>=n;i--)#define ll long long#define pa pair
           
            #define maxn 1000010#define inf 1000000000using namespace std;char s[maxn];int l,nxt[maxn];inline void getnext(){ int i=0,j=-1; nxt[0]=-1; while (i