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

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

這題的意思就是在文字串中找迴圈的串，輸出最後一個字元位置和迴圈節的長度

//e
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1e6+10;
char s[maxn];
int ne[maxn];
void get(int len)
{
	ne[0]=-1;
	int j=-1;
	for(int i=0;i<len;)
	{
		if(j==-1||s[i]==s[j]) ne[++i]=++j;
		else j=ne[j];
	}
}
int main()
{
	int t;
	int cas=1;
	while(~scanf("%d",&t)&&t)
	{
		scanf("%s",s);
		int len=strlen(s);
		get(len);
		
		printf("Test case #%d\n",cas++); 
		for(int i=2;i<=len;i++) 
		{
			if(ne[i]>0)
			{
				int n=i-ne[i];//判斷當前的（0到i）迴圈節長度 
				if(i%n==0) //中間沒有其他雜物  ！=0就不是迴圈節，直接continue； 
				 printf("%d %d\n",i,i/n);
			}
		}
		printf("\n");
	}
}