The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book: 

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais… 

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces. 

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap. 
 

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format: 

One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W). 
One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000. 

Output

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T. 
 

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Sample Output

1
3
0

//这题的意思就是第一个串在第二个串中的位置，kmp入门题，直接贴代码：

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1e6+10;
const int ma=1e4+10;
char a[maxn];
char b[ma];
int ne[maxn];
void get()  //常规处理方法 
{
	int len=strlen(b);
	ne[0]=-1;
	for(int i=0,j=-1;i<len;)
	{
		if(j==-1||b[i]==b[j]) ne[++i]=++j;
		else j=ne[j];
	}
}
int kmp()
{
	get();
	int n=strlen(a);
	int len=strlen(b);
	int ans=0;
	for(int i=0,j=0;i<n;)
	{
		if(j==-1||a[i]==b[j]) i++,j++;
		else j=ne[j];
		if(j>=len)  //当找到了就ans++统计次数 
		{
			ans++;
		}
	}
	return ans;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s",&b);
		scanf("%s",&a);
		printf("%d\n",kmp()); 
	}
}