#include <iostream>
#include <cstring>
using namespace std;

/*
 * 因为碱基对配对只有A T G C 四种基本碱基，极其容易出现重复序列，故采用kmp算法
 * 来解决问题
 */

char t[1000];//文本串
char p[1000];//模式串

int * nextBuild(const char *pattern)
{
    size_t m=strlen(pattern),j=0;
    int *N=new int[m];
    int state=N[0]=-1;
    while(j<m-1)
    {
        if(state<0|| pattern[j] == pattern[state])
        {

            state++,j++;
            N[j]= pattern[j] == pattern[state] ? N[state] : state;

        }else
        {
            state=N[state];
        }
    }
    return N;
}

void overturn(char *pattern)
{
    unsigned int i=0;
    while(pattern[i]!=0)
    {
        switch(pattern[i])
        {
            case 'A':
                pattern[i]='T';
                break;
            case 'T':
                pattern[i]='A';
                break;
            case 'G':
                pattern[i]='C';
                break;
            case 'C':
                pattern[i]='G';
                break;
        }
        i++;
    }
}
int match(char *text,char *pattern)
{
    int m=int(strlen(text)),i=0;
    int n=int(strlen(pattern)),j=0;
    int *next= nextBuild(pattern);
    while(i<m&&j<n)
    {
        if(j<0||text[i]==pattern[j])
        {
            j++,i++;
        }else
            j=next[j];
    }
    delete [] next;

    return i-j+1;
}

int main()
{
    gets(t);
    gets(p);
    overturn(p);
    int position= match(t,p);
    if(position<=(strlen(t)-strlen(p)+1))
    {
        printf("%d\n",position);
    }else
        cout<<0<<endl;
    return 0;
}