/*

Description

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

*/

/*

void GetNext(string pattern, int next[])

{

int patternSize = pattern.size();

int j, k;

next[0] = 0;

k = 0;

for( j = 1 ; j < patternSize; j++)

{

while( k > 0 && pattern[j] != pattern[k])

{

k = next[k];

}

if(pattern[j] == pattern[k])

{

k++;

}

next[j] = k;

}

}

*/

#include<cstring>

#include<iostream>

#include<cstdio>

using namespace std;

void GetNext(string pattern, int next[])

{

int patternSize = pattern.size();

next[0] = -1;

int k = -1;

int j = 0;

while( j < patternSize)

{

if( k == -1 || pattern[j] == pattern[k])

{

++k;

++j;

next[j] = k;

}

else

{

k = next[k];

}

}

}

int matchCount(string text, string pattern)

{

int textSize = text.size();

int patternSize = pattern.size();

int textIdx = 0; //模式串的0下标，在正文串中的位置

int patternIdx = 0; //模式串当前匹配到的索引

int* next = new int[patternSize+1];

GetNext(pattern,next);

int count = 0;

while(textIdx < textSize)

{

if(patternIdx <= patternSize)

{

if(text[textIdx] == pattern[patternIdx] || patternIdx == -1)

{

patternIdx++;

textIdx++;

if(patternIdx == patternSize)

{

count++;

}

}

else

{

patternIdx = next[patternIdx];

}

}

}

return count;

}

/*

int matchCount(string text, string pattern)

{

int textSize = text.size();

int patternSize = pattern.size();

int textIdx = 0; //模式串的0下标，在正文串中的位置

int patternIdx = 0; //模式串当前匹配到的索引

int count = 0;

while(textIdx < textSize)

{

if(patternIdx < patternSize)

{

if(text[textIdx + patternIdx] == pattern[patternIdx])

{

patternIdx++;

}

else

{

textIdx++;

patternIdx = 0;

}

}

else

{

count++;

patternIdx = 0;

textIdx++;

}

}

return count;

}

*/

int main(int argc, char** argv)

{

const char* INPUTFILE("../res/test-case/sample_input_strtest.txt");

freopen(INPUTFILE,"r",stdin);

int T;

cin >> T;

for(int i = 0 ; i < T; i++)

{

string pattern, text;

cin >> pattern;

cin >> text;

cout << matchCount(text,pattern) << endl;

}

}