链接：

 

Power Strings

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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.

 

代码：

#include
          
           #include
           
            #include
            
             #define N 1000007char S[N];int Next[N];  /// Next中存的是前缀和后缀的最大相似度void FindNext(int Slen, int Next[]) ///Next[i] 代表前 i 个字符的最大匹配度{    int i=0, j=-1;    Next[0] = -1;    while(i