#include <iostream>
#include <fstream>
#include <cmath>
#include <cstring>

using namespace std;

ifstream f("strmatch.in");
ofstream g("strmatch.out");

const int dim=2000001, mod1=pow(10, 9)+7, mod2=pow(10, 9)+9;

int main()
{
    int n, m, i, nr1=0, nr2=0, nr3=0, nr4=0, putere[dim], sol[dim], k=0;
    char a[dim], b[dim];
    f>> a >> b;
    n=strlen(a); m=strlen(b);
    putere[0]=1;
    nr1=a[0]-'A'; nr2=b[0]-'A';
    nr3=a[0]-'A'; nr4=b[0]-'A';
    for(i=1; i<n; i++){
        nr1=(nr1*26LL+a[i]-'A')%mod1; nr2=(nr2*26LL+b[i]-'A')%mod1;
        nr3=(nr3*26LL+a[i]-'A')%mod2; nr4=(nr4*26LL+b[i]-'A')%mod2;
        putere[i]=(putere[i-1]*26LL)%mod1;
    }
    putere[n]=(putere[n-1]*26LL)%mod1;
    if(nr1==nr2 && nr3==nr4){
        sol[++k]=0;
    }
    for(i=n; i<m; i++){
        nr2=((nr2*26LL+b[i]-'A')-(1LL*(b[i-n]-'A')*putere[n]%mod1)+mod1)%mod1;
        nr4=((nr4*26LL+b[i]-'A')-(1LL*(b[i-n]-'A')*putere[n]%mod2)+mod2)%mod2;
        if(nr1==nr2 && nr3==nr4){
            sol[++k]=i-n+1;
            if(k==1000){
                break;
            }
        }
    }
    g<< k << endl;
    for(i=1; i<=k; i++){
        g<< sol[i] << " ";
    }

    return 0;
}