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 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

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4
题解：如果一个长度为n的字符串(假设从0开始)，n%（n-nxt[n-1]）==0 ,那么n-nxt[n-1]就是这个字符串的最小循环长度，n/(n-nxt[n-1])为其最大循环周期，如果不等于0，设x=n%(n-nxt[n-1]) 那么该字符串加上x个字符后使得n-nxt[n-1] 为其最小循环长度
例题：http://acm.hdu.edu.cn/showproblem.php?pid=3746；

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e6+5;
typedef long long ll;
ll nxt[N]; 
struct stu{
    ll a,b;
    bool friend operator <(const stu &x,const stu &y){
        if(x.a!=y.a) return x.a<y.a;
    }
}arr[N];

int main(){
    int t,k=0;
    while(~scanf("%d",&t)!=EOF,t){
        memset(nxt,0,sizeof(nxt));
        k++;
        string a;
        cin>>a; 
        for(int i=1;i<a.size();i++){
            int j=nxt[i-1];
            while(j>0&&a[i]!=a[j]) j=nxt[j-1];
            if(a[i]==a[j]) j++;
            nxt[i]=j;
        } 
        ll pos=0; 
        printf("Test case #%d\n",k); 
        
        for(int i=1;i<a.size();i++) {
            int j=i+1;//总长度，
            // j%(j-nxt[i])用来判断j-nxt[i]是否可以作为该段的最小循环长度，
            // j/(j-nxt[i]) 循环周期 
            if(j%(j-nxt[i])==0 && j/(j-nxt[i])>=2){ 
                arr[pos].a=j;
                arr[pos++].b=j/(j-nxt[i]); 
            }
        } 
         sort(arr,arr+pos);
         for(int i=0;i<pos;i++){
             cout<<arr[i].a<<" "<<arr[i].b<<endl;
         }
         cout<<endl;
    }
    return 0;
}