Period

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

Output

For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3aaa12aabaabaabaab0

Sample Output

Test case #12 23 3Test case #22 26 29 312 4

求从1到n的串中，出现由循环节构成的串时输出他的位置和组成串的循环节的个数。其中当最小循环节是其本身时，也就是无循环节时（i==ans时）不输出

#include<stdio.h>#include<math.h>#include<string.h>#include<iostream>#include<algorithm>const int maxn=1e6+10;char s[maxn];int Next[maxn];void getNext(){    int i=0;    int j=-1;    int l=strlen(s);    while(i<l)    {        if(j==-1||s[i]==s[j])        {            i++;            j++;            Next[i]=j;        }        else j=Next[j];    }}int main(){    int n,r=1;    while(~scanf("%d",&n)&&n)    {        scanf("%s",s);        Next[0]=-1;        getNext();        printf("Test case #%d\n",r++);       for(int i=1;i<=n;i++)       {           int ans=i-Next[i];           if(i%ans==0&&i!=ans)           {               printf("%d %d\n",i,i/ans);           }       }       printf("\n");    }}