B -POJ2406 Hash

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.

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题意：输入若干个字符串，对于每个字符串求最短的循环节，求出他最多由几个相同的连续子串连接而成。

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#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef unsigned long long ull;
const int N = 1000000+10;
const int p = 131;
char str[N];
ull idx[N], hash[N];
int main()
{
    idx[0] = 1;
    //idx用来记录p的n次方；
    for(int i=1; i