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.

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.

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.

/*

Problem: 3461        User: bbsh

Memory: 5296K        Time: 110MS

Language: G++        Result: Accepted

*/

#include<cstdio>

#include<cstring>

using namespace std;

const int N=1e6+;

typedef int i64;

i64 S,chk,p,P,hash_key[N];

char s1[N],s2[N];

int cas,l1,l2,ans;

inline i64 fpow(i64 a,i64 p){

    i64 res=;

    for(;p;p>>=,a=a*a) if(p&) res=res*a;

    return res;

}

inline void get_s2_key(){

    p=fpow(P,l2);S=;

    for(int i=;i<=l2;i++) S=S*P+s2[i]-'A';

}

inline void get_s1_key(){

    hash_key[]=;

    for(int i=;i<=l1;i++) hash_key[i]=hash_key[i-]*P+s1[i]-'A';

}

inline i64 query(int x,int y){

    return hash_key[y]-hash_key[x-]*p;

}

int main(){

    P=;//prime_num

    for(scanf("%d",&cas);cas--;){

        scanf("%s%s",s2+,s1+);ans=;

        l1=strlen(s1+);l2=strlen(s2+);

        get_s2_key();

        get_s1_key();

        for(int i=l2;i<=l1;i++){

            chk=query(i-l2+,i);

            if(chk==S) ans++;

        }

        printf("%d\n",ans);

    }

    return ;

}

/*

Problem: 3461        User: bbsh

Memory: 1424K        Time: 110MS

Language: G++        Result: Accepted

Source Code

*/

#include<cstdio>

#include<cstring>

using namespace std;

const int N=1e6+;

char s1[N],s2[N];

int cas,l1,l2,fail[N];

void get_next(){

    int p=;fail[]=;

    for(int i=;i<=l2;i++){

        while(p>&&s2[i]!=s2[p+]) p=fail[p];

        if(s2[i]==s2[p+]) p++;

        fail[i]=p;

    }

}

void kmp(){

    int p=,ans=;

    for(int i=;i<=l1;i++){

        while(p>&&s1[i]!=s2[p+]) p=fail[p];

        if(s1[i]==s2[p+]) p++;

        if(p==l2) ans++,p=fail[p];

    }

    printf("%d\n",ans);

}

int main(){

    for(scanf("%d",&cas);cas--;){

        scanf("%s%s",s2+,s1+);

        l1=strlen(s1+);l2=strlen(s2+);

        get_next();

        kmp();

    }

    return ;

}